#set document(title: "12.4 Binomial Theorem", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 12.4#h(0.6em)Binomial Theorem #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Simplify: #math.equation(block: false, alt: "the fraction 7 times 6 times 5 times 4 over 4 times 3 times 2 times 1.")[$frac(7 · 6 · 5 · 4, 4 · 3 · 2 · 1) .$] #linebreak() If you missed this problem, review . #solutionbox[ 35 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Expand: #math.equation(block: false, alt: "open parenthesis 3 x plus 5 close parenthesis squared .")[$attach(( 3 x + 5 ), t: 2) .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "9 x squared plus 30 x plus 25")[$9 x^(2) + 30 x + 25$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Expand: #math.equation(block: false, alt: "open parenthesis x minus y close parenthesis squared .")[$attach(( x − y ), t: 2) .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "x squared minus 2 x y plus y squared")[$x^(2) − 2 x y + y^(2)$] ] ] === Use Pascal’s Triangle to Expand a Binomial In our previous work, we have squared binomials either by using FOIL or by using the Binomial Squares Pattern. We can also say that we expanded #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared .")[$attach(( a + b ), t: 2) .$] #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis squared equals a squared plus 2 a b plus b squared")[$attach(( a + b ), t: 2) = a^(2) + 2 a b + b^(2)$]To expand #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis cubed ,")[$attach(( a + b ), t: 3) ,$] we recognize that this is #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared open parenthesis a plus b close parenthesis")[$attach(( a + b ), t: 2) ( a + b )$] and multiply. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis cubed; open parenthesis a plus b close parenthesis squared open parenthesis a plus b close parenthesis; open parenthesis a squared plus 2 a b plus b squared close parenthesis open parenthesis a plus b close parenthesis; a cubed plus 2 a squared b plus a b squared plus a squared b plus 2 a b squared plus b cubed; a cubed plus 3 a squared b plus 3 a b squared plus b cubed; open parenthesis a plus b close parenthesis cubed equals a cubed plus 3 a squared b plus 3 a b squared plus b cubed")[$attach(( a + b ), t: 3) \ attach(( a + b ), t: 2) ( a + b ) \ ( a^(2) + 2 a b + b^(2) ) ( a + b ) \ a^(3) + 2 a^(2) b + a b^(2) + a^(2) b + 2 a b^(2) + b^(3) \ a^(3) + 3 a^(2) b + 3 a b^(2) + b^(3) \ attach(( a + b ), t: 3) = a^(3) + 3 a^(2) b + 3 a b^(2) + b^(3)$]To find a method that is less tedious that will work for higher expansions like #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 7 ,")[$attach(( a + b ), t: 7) ,$] we again look for patterns in some expansions. #figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Number of terms], [First term], [Last term]), [#math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 1 equals a plus b")[$attach(( a + b ), t: 1) = a + b$]], [2], [#math.equation(block: false, alt: "a to the power 1")[$a^(1)$]], [#math.equation(block: false, alt: "b to the power 1")[$b^(1)$]], [#math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared equals a squared plus 2 a b plus b squared")[$attach(( a + b ), t: 2) = a^(2) + 2 a b + b^(2)$]], [3], [#math.equation(block: false, alt: "a squared")[$a^(2)$]], [#math.equation(block: false, alt: "b squared")[$b^(2)$]], [#math.equation(block: false, alt: "open parenthesis a plus b close parenthesis cubed equals a cubed plus 3 a squared b plus 3 a b squared plus b cubed")[$attach(( a + b ), t: 3) = a^(3) + 3 a^(2) b + 3 a b^(2) + b^(3)$]], [4], [#math.equation(block: false, alt: "a cubed")[$a^(3)$]], [#math.equation(block: false, alt: "b cubed")[$b^(3)$]], [#math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 4 equals a to the power 4 plus 4 a cubed b plus 6 a squared b squared plus 4 a b cubed plus b to the power 4")[$attach(( a + b ), t: 4) = a^(4) + 4 a^(3) b + 6 a^(2) b^(2) + 4 a b^(3) + b^(4)$]], [5], [#math.equation(block: false, alt: "a to the power 4")[$a^(4)$]], [#math.equation(block: false, alt: "b to the power 4")[$b^(4)$]], [#math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 5 equals a to the power 5 plus 5 a to the power 4 b plus 10 a cubed b squared plus 10 a squared b cubed plus 5 a b to the power 4 plus b to the power 5")[$attach(( a + b ), t: 5) = a^(5) + 5 a^(4) b + 10 a^(3) b^(2) + 10 a^(2) b^(3) + 5 a b^(4) + b^(5)$]], [6], [#math.equation(block: false, alt: "a to the power 5")[$a^(5)$]], [#math.equation(block: false, alt: "b to the power 5")[$b^(5)$]], [#math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$attach(( a + b ), t: n)$]], [#math.equation(block: false, alt: "plus 1")[$+ 1$]], [#math.equation(block: false, alt: "a to the power n")[$a^(n)$]], [#math.equation(block: false, alt: "b to the power n")[$b^(n)$]], )) Notice the first and last terms show only one variable. Recall that #math.equation(block: false, alt: "a to the power 0 equals 1 ,")[$a^(0) = 1 ,$] so we could rewrite the first and last terms to include both variables. For example, we could expand #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis cubed")[$attach(( a + b ), t: 3)$] to show each term with both variables. #figure(figph[This figure shows the pattern a plus b to the power of 3 equals a to a power of 3 times b to a power of 0 plus 3 times a to a power of 2 times b to a power of 1 plus 3 a to a power of 0 times b to a power of 3.], alt: "This figure shows the pattern a plus b to the power of 3 equals a to a power of 3 times b to a power of 0 plus 3 times a to a power of 2 times b to a power of 1 plus 3 a to a power of 0 times b to a power of 3.", caption: none) Generally, we don’t show the zero exponents, just as we usually write #emph[x] rather than 1#emph[x]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Patterns in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$attach(( a + b ), t: n)$]] - The number of terms is #math.equation(block: false, alt: "n plus 1.")[$n + 1 .$] - The first term is #math.equation(block: false, alt: "a to the power n")[$a^(n)$] and the last term is #math.equation(block: false, alt: "b to the power n .")[$b^(n) .$] - The exponents on #emph[a] decrease by one on each term going left to right. - The exponents on #emph[b] increase by one on each term going left to right. - The sum of the exponents on any term is #emph[n]. ] Let’s look at an example to highlight the last three patterns. #figure(figph[This figure shows the pattern a plus b to the power of 5 equals a plus 5 times a times b plus 10 times a times b plus 5 times a times b plus b.], alt: "This figure shows the pattern a plus b to the power of 5 equals a plus 5 times a times b plus 10 times a times b plus 5 times a times b plus b.", caption: none) From the patterns we identified, we see the variables in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n ,")[$attach(( a + b ), t: n) ,$] would be #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals a to the power n plus ___ a to the power n minus 1 b to the power 1 plus ___ a to the power n minus 2 b squared plus ... plus ___ a to the power 1 b to the power n minus 1 plus b to the power n .")[$attach(( a + b ), t: n) = a^(n) + \_\_\_ a^(n − 1) b^(1) + \_\_\_ a^(n − 2) b^(2) + ... + \_\_\_ a^(1) b^(n − 1) + b^(n) .$]To find the coefficients of the terms, we write our expansions again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients. #figure(figph[A plus b to the power of 0 equals 1. The top level of Pascal’s Triangle is 1. A plus b to the power of 1 equals 1 a plus 1 b. The second level of Pascal’s Triangle is 1, 1. A plus b to the power of 2 equals 1 a to the power of 2 plus 2 a b plus 1 b to the power of 2. The third level of Pascal’s Triangle is 1, 2, 1. A plus b to the power of 3 equals 1 a to the power of 3 plus 3 a to the power of 2 b plus 3 a b to the power of 2 plus 1 b to the power of 3. The fourth level of Pascal’s Triangle is 1,3,3,1. A plus b to the power of 4 equals 1 a to the power of 4 plus 4 a to the power of 3 b plus 6 a to the power of 2 b to the power of 2 plus 4 a b to the power of 3 plus 1 b to the power of 4. The fifth level of Pascal’s Triangle is 1, 4, 6, 4, 1. A plus b to the power of 5 equals 1 a to the power of 5 plus 5 a to the power of 4 b plus 10 a to the power of 3 b to the power of 2 plus 10 a to the power of 2 b to the power of 3. The sixth row of the Pascal’s Triangle is 1, 5, 10, 10, 5, 1.], alt: "A plus b to the power of 0 equals 1. The top level of Pascal’s Triangle is 1. A plus b to the power of 1 equals 1 a plus 1 b. The second level of Pascal’s Triangle is 1, 1. A plus b to the power of 2 equals 1 a to the power of 2 plus 2 a b plus 1 b to the power of 2. The third level of Pascal’s Triangle is 1, 2, 1. A plus b to the power of 3 equals 1 a to the power of 3 plus 3 a to the power of 2 b plus 3 a b to the power of 2 plus 1 b to the power of 3. The fourth level of Pascal’s Triangle is 1,3,3,1. A plus b to the power of 4 equals 1 a to the power of 4 plus 4 a to the power of 3 b plus 6 a to the power of 2 b to the power of 2 plus 4 a b to the power of 3 plus 1 b to the power of 4. The fifth level of Pascal’s Triangle is 1, 4, 6, 4, 1. A plus b to the power of 5 equals 1 a to the power of 5 plus 5 a to the power of 4 b plus 10 a to the power of 3 b to the power of 2 plus 10 a to the power of 2 b to the power of 3. The sixth row of the Pascal’s Triangle is 1, 5, 10, 10, 5, 1.", caption: none) The array to the right is called #strong[Pascal’s Triangle]. Notice each number in the array is the sum of the two closest numbers in the row above. We can find the next row by starting and ending with one and then adding two adjacent numbers. #figure(figph[This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.], alt: "This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.", caption: none) This triangle gives the coefficients of the terms when we expand binomials. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Pascal’s Triangle] #figure(figph[This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.], alt: "This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.", caption: none) ] In the next example, we will use this triangle and the patterns we recognized to expand the binomial. #examplebox("Example 1")[][ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power 6.")[$attach(( x + y ), t: 6) .$] #solutionbox[ We know the variables for this expansion will follow the pattern we identified. The nonzero exponents of #emph[x] will start at six and decrease to one. The nonzero exponents of #emph[y] will start at one and increase to six. The sum of the exponents in each term will be six. In our pattern, #math.equation(block: false, alt: "a equals x")[$a = x$] and #math.equation(block: false, alt: "b equals y .")[$b = y .$] #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals a to the power n plus ___ a to the power n minus 1 b to the power 1 plus ___ a to the power n minus 2 b squared plus ... plus ___ a to the power 1 b to the power n minus 1 plus b to the power n; open parenthesis x plus y close parenthesis to the power 6 equals x to the power 6 plus ___ x to the power 5 y to the power 1 plus ___ x to the power 4 y squared plus ___ x cubed y cubed plus ___ x squared y to the power 4 plus ___ x to the power 1 y to the power 5 plus y to the power 6")[$attach(( a + b ), t: n) = a^(n) + \_\_\_ a^(n − 1) b^(1) + \_\_\_ a^(n − 2) b^(2) + ... + \_\_\_ a^(1) b^(n − 1) + b^(n) \ attach(( x + y ), t: 6) = x^(6) + \_\_\_ x^(5) y^(1) + \_\_\_ x^(4) y^(2) + \_\_\_ x^(3) y^(3) + \_\_\_ x^(2) y^(4) + \_\_\_ x^(1) y^(5) + y^(6)$]#figure(figph[This figure shows a plus b to the power of n equals a to the power of n plus a to the power if n minus 1 b to the power of 1 plus a to the power of n minus 2 b to the power if 2 plus ellipsis plus a to the power of 1 b to the power of n minus 1 plus b to the power of n. The next figure shows x plus y to the power of 6 equals x to the power of 6 plus x to the power of 5 y to the power of 1 plus x to the power of 4 y to the power of 2 plus x to the power of 3 y to the power of 3 plus x to the power of 2 y to the power of 4 plus x to the power of 1 y to the power of 5 plus y to the power of 6.], alt: "This figure shows a plus b to the power of n equals a to the power of n plus a to the power if n minus 1 b to the power of 1 plus a to the power of n minus 2 b to the power if 2 plus ellipsis plus a to the power of 1 b to the power of n minus 1 plus b to the power of n. The next figure shows x plus y to the power of 6 equals x to the power of 6 plus x to the power of 5 y to the power of 1 plus x to the power of 4 y to the power of 2 plus x to the power of 3 y to the power of 3 plus x to the power of 2 y to the power of 4 plus x to the power of 1 y to the power of 5 plus y to the power of 6.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power 5.")[$attach(( x + y ), t: 5) .$] #solutionbox[ #math.equation(block: true, alt: "x to the power 5 plus 5 x to the power 4 y plus 10 x cubed y squared plus 10 x squared y cubed")[$x^(5) + 5 x^(4) y + 10 x^(3) y^(2) + 10 x^(2) y^(3)$] #linebreak() #math.equation(block: true, alt: "plus 5 x y to the power 4 plus y to the power 5")[$+ 5 x y^(4) + y^(5)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis p plus q close parenthesis to the power 7.")[$attach(( p + q ), t: 7) .$] #solutionbox[ #math.equation(block: true, alt: "p to the power 7 plus 7 p to the power 6 q plus 21 p to the power 5 q squared plus 35 p to the power 4 q cubed")[$p^(7) + 7 p^(6) q + 21 p^(5) q^(2) + 35 p^(4) q^(3)$] #linebreak() #math.equation(block: true, alt: "plus 35 p cubed q to the power 4 plus 21 p squared q to the power 5 plus 7 p q to the power 6 plus q to the power 7")[$+ 35 p^(3) q^(4) + 21 p^(2) q^(5) + 7 p q^(6) + q^(7)$] ] ] In the next example we want to expand a binomial with one variable and one constant. We need to identify the #emph[a] and #emph[b] to carefully apply the pattern. #examplebox("Example 2")[][ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis x plus 3 close parenthesis to the power 5.")[$attach(( x + 3 ), t: 5) .$] #solutionbox[ We identify the #emph[a] and #emph[b] of the pattern. #figure(figph[This figure shows how we identify a plus b to the power of n, in the pattern x plus 3 to the power of 5.], alt: "This figure shows how we identify a plus b to the power of n, in the pattern x plus 3 to the power of 5.", caption: none) In our pattern, #math.equation(block: false, alt: "a equals x")[$a = x$] and #math.equation(block: false, alt: "b equals 3.")[$b = 3 .$] We know the variables for this expansion will follow the pattern we identified. The sum of the exponents in each term will be five. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals a to the power n plus ___ a to the power n minus 1 b to the power 1 plus ___ a to the power n minus 2 b squared plus ... plus ___ a to the power 1 b to the power n minus 1 plus b to the power n; open parenthesis x plus 3 close parenthesis to the power 5 equals x to the power 5 plus ___ x to the power 4 times 3 to the power 1 plus ___ x cubed times 3 squared plus ___ x squared times 3 cubed plus ___ x to the power 1 times 3 to the power 4 plus 3 to the power 5")[$attach(( a + b ), t: n) = a^(n) + \_\_\_ a^(n − 1) b^(1) + \_\_\_ a^(n − 2) b^(2) + ... + \_\_\_ a^(1) b^(n − 1) + b^(n) \ attach(( x + 3 ), t: 5) = x^(5) + \_\_\_ x^(4) · 3^(1) + \_\_\_ x^(3) · 3^(2) + \_\_\_ x^(2) · 3^(3) + \_\_\_ x^(1) · 3^(4) + 3^(5)$]#figure(figph[This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1. This figure shows X plus 3 to the power of 5 equals 1 x to the power of 5 g 3 x to the power of 4 plus 10 g 9 x to the power of 3 plus 10 g 27 x to the power of 2 plus 5 g 81 x to the power of 1 plus 1 g 243. Then, x plus 3 to the power of 5 equals x to the power of 5 plus 15 x to the power of 4 plus 90 x to the power of 3 plus 270 x to the power of 2 plus 405 plus 243.], alt: "This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1. This figure shows X plus 3 to the power of 5 equals 1 x to the power of 5 g 3 x to the power of 4 plus 10 g 9 x to the power of 3 plus 10 g 27 x to the power of 2 plus 5 g 81 x to the power of 1 plus 1 g 243. Then, x plus 3 to the power of 5 equals x to the power of 5 plus 15 x to the power of 4 plus 90 x to the power of 3 plus 270 x to the power of 2 plus 405 plus 243.", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis to the power 4.")[$attach(( x + 2 ), t: 4) .$] #solutionbox[ #math.equation(block: true, alt: "x to the power 4 plus 8 x cubed plus 24 x squared plus 32 x plus 16")[$x^(4) + 8 x^(3) + 24 x^(2) + 32 x + 16$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis x plus 1 close parenthesis to the power 6.")[$attach(( x + 1 ), t: 6) .$] #solutionbox[ #math.equation(block: true, alt: "x to the power 6 plus 6 x to the power 5 plus 15 x to the power 4 plus 20 x cubed plus 15 x squared")[$x^(6) + 6 x^(5) + 15 x^(4) + 20 x^(3) + 15 x^(2)$] #linebreak() #math.equation(block: true, alt: "plus 6 x plus 1")[$+ 6 x + 1$] ] ] In the next example, the binomial is a difference and the first term has a constant times the variable. Once we identify the #emph[a] and #emph[b] of the pattern, we must once again carefully apply the pattern. #examplebox("Example 3")[][ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis 3 x minus 2 close parenthesis to the power 4.")[$attach(( 3 x − 2 ), t: 4) .$] #solutionbox[ We identify the #emph[a] and #emph[b] of the pattern. #figure(figph[This figure shows how we identify a plus b to the power of n, in the pattern 3 x minus 2 to the power of 4.], alt: "This figure shows how we identify a plus b to the power of n, in the pattern 3 x minus 2 to the power of 4.", caption: none) In our pattern, #math.equation(block: false, alt: "a equals 3 x")[$a = 3 x$] and #math.equation(block: false, alt: "b equals −2.")[$b = −2 .$] #figure(figph[This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.], alt: "This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.", caption: none) #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals a to the power n plus ___ a to the power n minus 1 b to the power 1 plus ___ a to the power n minus 2 b squared plus ... plus ___ a to the power 1 b to the power n minus 1 plus b to the power n; open parenthesis 3 x minus 2 close parenthesis to the power 4 equals 1 times open parenthesis 3 x close parenthesis to the power 4 plus 4 open parenthesis 3 x close parenthesis cubed open parenthesis −2 close parenthesis to the power 1 plus 6 open parenthesis 3 x close parenthesis squared open parenthesis −2 close parenthesis squared plus 4 open parenthesis 3 x close parenthesis to the power 1 open parenthesis −2 close parenthesis cubed plus 1 times open parenthesis −2 close parenthesis to the power 4; open parenthesis 3 x minus 2 cl")[$attach(( a + b ), t: n) = a^(n) + \_\_\_ a^(n − 1) b^(1) + \_\_\_ a^(n − 2) b^(2) + ... + \_\_\_ a^(1) b^(n − 1) + b^(n) \ attach(( 3 x − 2 ), t: 4) = 1 · attach(( 3 x ), t: 4) + 4 attach(( 3 x ), t: 3) attach(( −2 ), t: 1) + 6 attach(( 3 x ), t: 2) attach(( −2 ), t: 2) + 4 attach(( 3 x ), t: 1) attach(( −2 ), t: 3) + 1 · attach(( −2 ), t: 4) \ attach(( 3 x − 2 ), t: 4) = 81 x^(4) + 4 ( 27 x^(3) ) ( −2 ) + 6 ( 9 x^(2) ) ( 4 ) + 4 ( 3 x ) ( −8 ) + 1 · 16 \ attach(( 3 x − 2 ), t: 4) = 81 x^(4) − 216 x^(3) + 216 x^(2) − 96 x + 16$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis 2 x minus 3 close parenthesis to the power 4.")[$attach(( 2 x − 3 ), t: 4) .$] #solutionbox[ #math.equation(block: true, alt: "16 x to the power 4 minus 96 x cubed plus 216 x squared minus 216 x plus 81")[$16 x^(4) − 96 x^(3) + 216 x^(2) − 216 x + 81$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use Pascal’s Triangle to expand #math.equation(block: false, alt: "open parenthesis 2 x minus 1 close parenthesis to the power 6.")[$attach(( 2 x − 1 ), t: 6) .$] #solutionbox[ #math.equation(block: true, alt: "64 x to the power 6 minus 192 x to the power 5 plus 240 x to the power 4 minus 160 x cubed")[$64 x^(6) − 192 x^(5) + 240 x^(4) − 160 x^(3)$] #linebreak() #math.equation(block: true, alt: "plus 60 x squared minus 12 x plus 1")[$+ 60 x^(2) − 12 x + 1$] ] ] === Evaluate a Binomial Coefficient While Pascal’s Triangle is one method to expand a binomial, we will also look at another method. Before we get to that, we need to introduce some more #strong[factorial notation]. This notation is not only used to expand binomials, but also in the study and use of probability. To find the coefficients of the terms of expanded binomials, we will need to be able to evaluate the notation #math.equation(block: false, alt: "open parenthesis n; r close parenthesis")[$( n \ r )$] which is called a #strong[binomial coefficient]. We read #math.equation(block: false, alt: "open parenthesis n; r close parenthesis")[$( n \ r )$] as “#emph[n] choose #emph[r]” or “#emph[n] taken #emph[r] at a time”. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Binomial Coefficient #math.equation(block: false, alt: "open parenthesis n; r close parenthesis")[$( n \ r )$]] A binomial coefficient #math.equation(block: false, alt: "open parenthesis n; r close parenthesis ,")[$( n \ r ) ,$] where #emph[r] and #emph[n] are integers with #math.equation(block: false, alt: "0 less than or equal to r less than or equal to n ,")[$0 ≤ r ≤ n ,$] is defined as #math.equation(block: true, alt: "open parenthesis n; r close parenthesis equals the fraction n ! over r ! open parenthesis n minus r close parenthesis !")[$( n \ r ) = frac(n !, r ! ( n − r ) !)$]We read #math.equation(block: false, alt: "open parenthesis n; r close parenthesis")[$( n \ r )$] as “#emph[n] choose #emph[r]” or “#emph[n] taken #emph[r] at a time”. ] #examplebox("Example 4")[][ Evaluate: ⓐ #math.equation(block: false, alt: "open parenthesis 5; 1 close parenthesis")[$( 5 \ 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 7; 7 close parenthesis")[$( 7 \ 7 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 4; 0 close parenthesis")[$( 4 \ 0 )$] ⓓ #math.equation(block: false, alt: "open parenthesis 8; 5 close parenthesis .")[$( 8 \ 5 ) .$] #solutionbox[ ⓐ We will use the definition of a binomial coefficient, #math.equation(block: false, alt: "open parenthesis n; r close parenthesis equals the fraction n ! over r ! open parenthesis n minus r close parenthesis ! .")[$( n \ r ) = frac(n !, r ! ( n − r ) !) .$] #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "open parenthesis 5; 1 close parenthesis")[$#h(4em) ( 5 \ 1 )$]]), [Use the definition, #math.equation(block: false, alt: "open parenthesis n; r close parenthesis equals the fraction n ! over r ! open parenthesis n minus r close parenthesis ! ,")[$( n \ r ) = frac(n !, r ! ( n − r ) !) ,$] where #linebreak() #math.equation(block: false, alt: "n equals 5 , r equals 1.")[$n = 5 , r = 1 .$]], [#math.equation(block: false, alt: "the fraction 5 ! over 1 ! open parenthesis 5 minus 1 close parenthesis !")[$#h(4em) frac(5 !, 1 ! ( 5 − 1 ) !)$]], [Simplify.], [#math.equation(block: false, alt: "the fraction 5 ! over 1 ! open parenthesis 4 close parenthesis !")[$#h(4em) frac(5 !, 1 ! ( 4 ) !)$]], [Rewrite #math.equation(block: false, alt: "5 ! as 5 times 4 !")[$5 ! #h(0.2em) "as" #h(0.2em) 5 · 4 !$]], [#math.equation(block: false, alt: "the fraction 5 times 4 ! over 1 ! times 4 !")[$#h(4em) frac(5 · 4 !, 1 ! · 4 !)$]], [Simplify, by removing common factors.], [#math.equation(block: false, alt: "the fraction 5 times 4 ! over 1 times 4 !")[$#h(4em) frac(5 · cancel(4 !), 1 · cancel(4 !))$]], [Simplify.], [#math.equation(block: false, alt: "5")[$#h(4em) 5$]], [], [#math.equation(block: false, alt: "open parenthesis 5; 1 close parenthesis equals 5")[$#h(4em) ( 5 \ 1 ) = 5$]], )) ⓑ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "open parenthesis 7; 7 close parenthesis")[$#h(4em) ( 7 \ 7 )$]]), [Use the definition, #math.equation(block: false, alt: "open parenthesis n; r close parenthesis equals the fraction n ! over r ! open parenthesis n minus r close parenthesis ! ,")[$( n \ r ) = frac(n !, r ! ( n − r ) !) ,$] where #linebreak() #math.equation(block: false, alt: "n equals 7 , r equals 7.")[$n = 7 , r = 7 .$]], [#math.equation(block: false, alt: "the fraction 7 ! over 7 ! open parenthesis 7 minus 7 close parenthesis !")[$#h(4em) frac(7 !, 7 ! ( 7 − 7 ) !)$]], [Simplify.], [#math.equation(block: false, alt: "the fraction 7 ! over 7 ! open parenthesis 0 close parenthesis !")[$#h(4em) frac(7 !, 7 ! ( 0 ) !)$]], [Simplify. Remember #math.equation(block: false, alt: "0 ! equals 1.")[$0 ! = 1 .$]], [#math.equation(block: false, alt: "1")[$#h(4em) 1$]], [], [#math.equation(block: false, alt: "open parenthesis 7; 7 close parenthesis equals 1")[$#h(4em) ( 7 \ 7 ) = 1$]], )) ⓒ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "open parenthesis 4; 0 close parenthesis")[$#h(4em) ( 4 \ 0 )$]]), [Use the definition, #math.equation(block: false, alt: "open parenthesis n; r close parenthesis equals the fraction n ! over r ! open parenthesis n minus r close parenthesis ! ,")[$( n \ r ) = frac(n !, r ! ( n − r ) !) ,$] where #linebreak() #math.equation(block: false, alt: "n equals 4 , r equals 0.")[$n = 4 , r = 0 .$]], [#math.equation(block: false, alt: "the fraction 4 ! over 0 ! open parenthesis 4 minus 0 close parenthesis !")[$#h(4em) frac(4 !, 0 ! ( 4 − 0 ) !)$]], [Simplify.], [#math.equation(block: false, alt: "the fraction 4 ! over 0 ! open parenthesis 4 close parenthesis !")[$#h(4em) frac(4 !, 0 ! ( 4 ) !)$]], [Simplify.], [#math.equation(block: false, alt: "1")[$#h(4em) 1$]], [], [#math.equation(block: false, alt: "open parenthesis 4; 0 close parenthesis equals 1")[$#h(4em) ( 4 \ 0 ) = 1$]], )) ⓓ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "open parenthesis 8; 5 close parenthesis")[$#h(1.5em) ( 8 \ 5 )$]]), [Use the definition, #math.equation(block: false, alt: "open parenthesis n; r close parenthesis equals the fraction n ! over r ! open parenthesis n minus r close parenthesis ! ,")[$( n \ r ) = frac(n !, r ! ( n − r ) !) ,$] where #linebreak() #math.equation(block: false, alt: "n equals 8 , r equals 5.")[$n = 8 , r = 5 .$]], [#math.equation(block: false, alt: "the fraction 8 ! over 5 ! open parenthesis 8 minus 5 close parenthesis !")[$#h(1.5em) frac(8 !, 5 ! ( 8 − 5 ) !)$]], [Simplify.], [#math.equation(block: false, alt: "the fraction 8 ! over 5 ! open parenthesis 3 close parenthesis !")[$#h(1.5em) frac(8 !, 5 ! ( 3 ) !)$]], [Rewrite #math.equation(block: false, alt: "8 ! as 8 times 7 times 6 times 5 !")[$8 ! #h(0.2em) "as" #h(0.2em) 8 · 7 · 6 · 5 !$] and remove common factors.], [#math.equation(block: false, alt: "the fraction 8 times 7 times 6 times 5 ! over 5 ! times 3 times 2 times 1")[$#h(1.5em) frac(8 · 7 · cancel(6) · cancel(5 !), cancel(5 !) · cancel(3) · cancel(2) · 1)$]], [Simplify.], [#math.equation(block: false, alt: "56")[$#h(1.5em) 56$]], [], [#math.equation(block: false, alt: "open parenthesis 8; 5 close parenthesis equals 56")[$#h(1.5em) ( 8 \ 5 ) = 56$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Evaluate each binomial coefficient: ⓐ #math.equation(block: false, alt: "open parenthesis 6; 1 close parenthesis")[$( 6 \ 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 8; 8 close parenthesis")[$( 8 \ 8 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 5; 0 close parenthesis")[$( 5 \ 0 )$] ⓓ #math.equation(block: false, alt: "open parenthesis 7; 3 close parenthesis .")[$( 7 \ 3 ) .$] #solutionbox[ ⓐ 6 ⓑ 1 ⓒ 1 ⓓ 35 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Evaluate each binomial coefficient: ⓐ #math.equation(block: false, alt: "open parenthesis 2; 1 close parenthesis")[$( 2 \ 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 11; 11 close parenthesis")[$( 11 \ 11 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 9; 0 close parenthesis")[$( 9 \ 0 )$] ⓓ #math.equation(block: false, alt: "open parenthesis 6; 5 close parenthesis .")[$( 6 \ 5 ) .$] #solutionbox[ ⓐ 2 ⓑ 1 ⓒ 1 ⓓ 6 ] ] In the previous example, parts (a), (b), (c) demonstrate some special properties of binomial coefficients. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Properties of Binomial Coefficients] #math.equation(block: true, alt: "open parenthesis n; 1 close parenthesis equals n open parenthesis n; n close parenthesis equals 1 open parenthesis n; 0 close parenthesis equals 1")[$( n \ 1 ) = n #h(4em) ( n \ n ) = 1 #h(4em) ( n \ 0 ) = 1$] ] === Use the Binomial Theorem to Expand a Binomial We are now ready to use the alternate method of expanding binomials. The #strong[Binomial Theorem] uses the same pattern for the variables, but uses the binomial coefficient for the coefficient of each term. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Binomial Theorem] For any real numbers #emph[a] and #emph[b], and positive integer #emph[n], #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals open parenthesis n; 0 close parenthesis a to the power n plus open parenthesis n; 1 close parenthesis a to the power n minus 1 b to the power 1 plus open parenthesis n; 2 close parenthesis a to the power n minus 2 b squared plus ... plus open parenthesis n; r close parenthesis a to the power n minus r b to the power r plus ... plus open parenthesis n; n close parenthesis b to the power n")[$attach(( a + b ), t: n) = ( n \ 0 ) a^(n) + ( n \ 1 ) a^(n − 1) b^(1) + ( n \ 2 ) a^(n − 2) b^(2) + ... + ( n \ r ) a^(n − r) b^(r) + ... + ( n \ n ) b^(n)$] ] #examplebox("Example 5")[][ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis p plus q close parenthesis to the power 4.")[$attach(( p + q ), t: 4) .$] #solutionbox[ We identify the #emph[a] and #emph[b] of the pattern. #figure(figph[This figure shows how we identify a plus b to the power of n, in the pattern p plus q to the power of 4.], alt: "This figure shows how we identify a plus b to the power of n, in the pattern p plus q to the power of 4.", caption: none) In our pattern, #math.equation(block: false, alt: "a equals p")[$a = p$] and #math.equation(block: false, alt: "b equals q .")[$b = q .$] We use the Binomial Theorem. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals open parenthesis n; 0 close parenthesis a to the power n plus open parenthesis n; 1 close parenthesis a to the power n minus 1 b to the power 1 plus open parenthesis n; 2 close parenthesis a to the power n minus 2 b squared plus ... plus open parenthesis n; r close parenthesis a to the power n minus r b to the power r plus ... plus open parenthesis n; n close parenthesis b to the power n")[$attach(( a + b ), t: n) = ( n \ 0 ) a^(n) + ( n \ 1 ) a^(n − 1) b^(1) + ( n \ 2 ) a^(n − 2) b^(2) + ... + ( n \ r ) a^(n − r) b^(r) + ... + ( n \ n ) b^(n)$]Substitute in the values #math.equation(block: false, alt: "a equals p ,")[$a = p ,$]#math.equation(block: false, alt: "b equals q")[$b = q$] and #math.equation(block: false, alt: "n equals 4.")[$n = 4 .$] #math.equation(block: true, alt: "open parenthesis p plus q close parenthesis to the power 4 equals open parenthesis 4; 0 close parenthesis p to the power 4 plus open parenthesis 4; 1 close parenthesis p to the power 4 minus 1 q to the power 1 plus open parenthesis 4; 2 close parenthesis p to the power 4 minus 2 q squared plus open parenthesis 4; 3 close parenthesis p to the power 4 minus 3 q cubed plus open parenthesis 4; 4 close parenthesis q to the power 4")[$attach(( p + q ), t: 4) = ( 4 \ 0 ) p^(4) + ( 4 \ 1 ) p^(4 − 1) q^(1) + ( 4 \ 2 ) p^(4 − 2) q^(2) + ( 4 \ 3 ) p^(4 − 3) q^(3) + ( 4 \ 4 ) q^(4)$]Simplify the exponents. #math.equation(block: true, alt: "open parenthesis p plus q close parenthesis to the power 4 equals open parenthesis 4; 0 close parenthesis p to the power 4 plus open parenthesis 4; 1 close parenthesis p cubed q plus open parenthesis 4; 2 close parenthesis p squared q squared plus open parenthesis 4; 3 close parenthesis p q cubed plus open parenthesis 4; 4 close parenthesis q to the power 4")[$attach(( p + q ), t: 4) = ( 4 \ 0 ) p^(4) + ( 4 \ 1 ) p^(3) q^ + ( 4 \ 2 ) p^(2) q^(2) + ( 4 \ 3 ) p^ q^(3) + ( 4 \ 4 ) q^(4)$]Evaluate the coefficients. Remember, #math.equation(block: false, alt: "open parenthesis n; 1 close parenthesis equals n ,")[$( n \ 1 ) = n ,$]#math.equation(block: false, alt: "open parenthesis n; n close parenthesis equals 1 ,")[$( n \ n ) = 1 ,$]#math.equation(block: false, alt: "open parenthesis n; 0 close parenthesis equals 1.")[$( n \ 0 ) = 1 .$] #math.equation(block: true, alt: "open parenthesis p plus q close parenthesis to the power 4 equals 1 p to the power 4 plus 4 p cubed q to the power 1 plus the fraction 4 ! over 2 ! open parenthesis 2 close parenthesis ! p squared q squared plus the fraction 4 ! over 3 ! open parenthesis 4 minus 3 close parenthesis ! p to the power 1 q cubed plus 1 q to the power 4; open parenthesis p plus q close parenthesis to the power 4 equals p to the power 4 plus 4 p cubed q plus 6 p squared q squared plus 4 p q cubed plus q to the power 4")[$attach(( p + q ), t: 4) = 1 p^(4) + 4 p^(3) q^(1) + frac(4 !, 2 ! ( 2 ) !) p^(2) q^(2) + frac(4 !, 3 ! ( 4 − 3 ) !) p^(1) q^(3) + 1 q^(4) \ attach(( p + q ), t: 4) = p^(4) + 4 p^(3) q^ + 6 p^(2) q^(2) + 4 p^ q^(3) + q^(4)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power 5.")[$attach(( x + y ), t: 5) .$] #solutionbox[ #math.equation(block: true, alt: "x to the power 5 plus 5 x to the power 4 y plus 10 x cubed y squared plus 10 x squared y cubed")[$x^(5) + 5 x^(4) y + 10 x^(3) y^(2) + 10 x^(2) y^(3)$] #linebreak() #math.equation(block: true, alt: "plus 5 x y to the power 4 plus y to the power 5")[$+ 5 x y^(4) + y^(5)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis m plus n close parenthesis to the power 6.")[$attach(( m + n ), t: 6) .$] #solutionbox[ #math.equation(block: true, alt: "m to the power 6 plus 6 m to the power 5 n plus 15 m to the power 4 n squared plus 20 m cubed n cubed")[$m^(6) + 6 m^(5) n + 15 m^(4) n^(2) + 20 m^(3) n^(3)$] #linebreak() #math.equation(block: true, alt: "plus 15 m squared n to the power 4 plus 6 m n to the power 5 plus n to the power 6")[$+ 15 m^(2) n^(4) + 6 m n^(5) + n^(6)$] ] ] Notice that when we expanded #math.equation(block: false, alt: "open parenthesis p plus q close parenthesis to the power 4")[$attach(( p + q ), t: 4)$] in the last example, using the Binomial Theorem, we got the same coefficients we would get from using #strong[Pascal’s Triangle]. #figure(figph[The figure above is P plus q to the power of 4 equals 4 choose 0 times p to the power of 4 plus 4 choose 1 times p to the power of 3 q plus 4 choose 2 times p to the power of 2 q to the power of 2 plus 4 choose 3 times p q to the power of 3 plus 4 choose 4 times q to the power of 4. P plus q to the power of 4 equals p to the power of 4 p to the power of 3 q plus 6 p to the power of 2 q to the power of 2 plus 4 p q to the power of 3 plus q to the power of 4. This figure on the right shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.], alt: "The figure above is P plus q to the power of 4 equals 4 choose 0 times p to the power of 4 plus 4 choose 1 times p to the power of 3 q plus 4 choose 2 times p to the power of 2 q to the power of 2 plus 4 choose 3 times p q to the power of 3 plus 4 choose 4 times q to the power of 4. P plus q to the power of 4 equals p to the power of 4 p to the power of 3 q plus 6 p to the power of 2 q to the power of 2 plus 4 p q to the power of 3 plus q to the power of 4. This figure on the right shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.", caption: none) The next example, the binomial is a difference. When the binomial is a difference, we must be careful in identifying the values we will use in the pattern. #examplebox("Example 6")[][ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis to the power 5.")[$attach(( x − 2 ), t: 5) .$] #solutionbox[ We identify the #emph[a] and #emph[b] of the pattern. #figure(figph[This figure shows x minus 2 to the power of 5.], alt: "This figure shows x minus 2 to the power of 5.", caption: none) In our pattern, #math.equation(block: false, alt: "a equals x")[$a = x$] and #math.equation(block: false, alt: "b equals −2.")[$b = −2 .$] We use the Binomial Theorem. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals open parenthesis n; 0 close parenthesis a to the power n plus open parenthesis n; 1 close parenthesis a to the power n minus 1 b to the power 1 plus open parenthesis n; 2 close parenthesis a to the power n minus 2 b squared plus ... plus open parenthesis n; r close parenthesis a to the power n minus r b to the power r plus ... plus open parenthesis n; n close parenthesis b to the power n")[$attach(( a + b ), t: n) = ( n \ 0 ) a^(n) + ( n \ 1 ) a^(n − 1) b^(1) + ( n \ 2 ) a^(n − 2) b^(2) + ... + ( n \ r ) a^(n − r) b^(r) + ... + ( n \ n ) b^(n)$]Substitute in the values #math.equation(block: false, alt: "a equals x ,")[$a = x ,$]#math.equation(block: false, alt: "b equals −2 ,")[$b = −2 ,$] and #math.equation(block: false, alt: "n equals 5.")[$n = 5 .$] #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis to the power 5 equals open parenthesis 5; 0 close parenthesis x to the power 5 plus open parenthesis 5; 1 close parenthesis x to the power 5 minus 1 open parenthesis −2 close parenthesis to the power 1 plus open parenthesis 5; 2 close parenthesis x to the power 5 minus 2 open parenthesis −2 close parenthesis squared plus open parenthesis 5; 3 close parenthesis x to the power 5 minus 3 open parenthesis −2 close parenthesis cubed plus open parenthesis 5; 4 close parenthesis x to the power 5 minus 4 open parenthesis −2 close parenthesis to the power 4 plus open parenthesis 5; 5 close parenthesis open parenthesis −2 close parenthesis to the power 5")[$attach(( x − 2 ), t: 5) = ( 5 \ 0 ) x^(5) + ( 5 \ 1 ) x^(5 − 1) attach(( −2 ), t: 1) + ( 5 \ 2 ) x^(5 − 2) attach(( −2 ), t: 2) + ( 5 \ 3 ) x^(5 − 3) attach(( −2 ), t: 3) + ( 5 \ 4 ) x^(5 − 4) attach(( −2 ), t: 4) + ( 5 \ 5 ) attach(( −2 ), t: 5)$]Simplify the exponents and evaluate the coefficients. Remember,#math.equation(block: false, alt: "open parenthesis n; 1 close parenthesis equals n ,")[$( n \ 1 ) = n ,$]#math.equation(block: false, alt: "open parenthesis n; n close parenthesis equals 1 ,")[$( n \ n ) = 1 ,$]#math.equation(block: false, alt: "open parenthesis n; 0 close parenthesis equals 1.")[$( n \ 0 ) = 1 .$] #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis to the power 5 equals open parenthesis 5; 0 close parenthesis x to the power 5 plus open parenthesis 5; 1 close parenthesis x to the power 4 open parenthesis −2 close parenthesis plus open parenthesis 5; 2 close parenthesis x cubed open parenthesis −2 close parenthesis squared plus open parenthesis 5; 3 close parenthesis x squared open parenthesis −2 close parenthesis cubed plus open parenthesis 5; 4 close parenthesis x open parenthesis −2 close parenthesis to the power 4 plus open parenthesis 5; 5 close parenthesis open parenthesis −2 close parenthesis to the power 5; open parenthesis x minus 2 close parenthesis to the power 5 equals 1 x to the power 5 plus 5 open parenthesis −2 close parenthesis x to the power 4 plus the fraction 5 ! over 2 !")[$attach(( x − 2 ), t: 5) = ( 5 \ 0 ) x^(5) + ( 5 \ 1 ) x^(4) attach(( −2 )) + ( 5 \ 2 ) x^(3) attach(( −2 ), t: 2) + ( 5 \ 3 ) x^(2) attach(( −2 ), t: 3) + ( 5 \ 4 ) x attach(( −2 ), t: 4) + ( 5 \ 5 ) attach(( −2 ), t: 5) \ attach(( x − 2 ), t: 5) = 1 x^(5) + 5 ( −2 ) x^(4) + frac(5 !, 2 ! · 3 !) attach(( −2 ), t: 2) x^(3) + frac(5 !, 3 ! · 2 !) attach(( −2 ), t: 3) x^(2) + frac(5 !, 4 ! · 1 !) attach(( −2 ), t: 4) x + 1 attach(( −2 ), t: 5) \ attach(( x − 2 ), t: 5) = x^(5) + 5 ( −2 ) x^(4) + 10 · 4 · x^(3) + 10 ( −8 ) x^(2) + 5 · 16 · x + 1 ( −32 ) \ attach(( x − 2 ), t: 5) = x^(5) − 10 x^(4) + 40 x^(3) − 80 x^(2) + 80 x − 32$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis to the power 5.")[$attach(( x − 3 ), t: 5) .$] #solutionbox[ #math.equation(block: true, alt: "x to the power 5 minus 15 x to the power 4 plus 90 x cubed minus 270 x squared")[$x^(5) − 15 x^(4) + 90 x^(3) − 270 x^(2)$] #linebreak() #math.equation(block: true, alt: "plus 405 x minus 243")[$+ 405 x − 243$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis y minus 1 close parenthesis to the power 6.")[$attach(( y − 1 ), t: 6) .$] #solutionbox[ #math.equation(block: true, alt: "y to the power 6 minus 6 y to the power 5 plus 15 y to the power 4 minus 20 y cubed plus 15 y squared")[$y^(6) − 6 y^(5) + 15 y^(4) − 20 y^(3) + 15 y^(2)$] #linebreak() #math.equation(block: true, alt: "minus 6 y plus 1")[$− 6 y + 1$] ] ] Things can get messy when both terms have a coefficient and a variable. #examplebox("Example 7")[][ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis 2 x minus 3 y close parenthesis to the power 4.")[$attach(( 2 x − 3 y ), t: 4) .$] #solutionbox[ We identify the #emph[a] and #emph[b] of the pattern. #figure(figph[This figure shows how we identify a plus b to the power of n, in the pattern 2 x minus 3 y times the power of 4.], alt: "This figure shows how we identify a plus b to the power of n, in the pattern 2 x minus 3 y times the power of 4.", caption: none) In our pattern, #math.equation(block: false, alt: "a equals 2 x")[$a = 2 x$] and #math.equation(block: false, alt: "b equals −3 y .")[$b = −3 y .$] We use the Binomial Theorem. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals open parenthesis n; 0 close parenthesis a to the power n plus open parenthesis n; 1 close parenthesis a to the power n minus 1 b to the power 1 plus open parenthesis n; 2 close parenthesis a to the power n minus 2 b squared plus ... plus open parenthesis n; r close parenthesis a to the power n minus r b to the power r plus ... plus open parenthesis n; n close parenthesis b to the power n")[$attach(( a + b ), t: n) = ( n \ 0 ) a^(n) + ( n \ 1 ) a^(n − 1) b^(1) + ( n \ 2 ) a^(n − 2) b^(2) + ... + ( n \ r ) a^(n − r) b^(r) + ... + ( n \ n ) b^(n)$]Substitute in the values #math.equation(block: false, alt: "a equals 2 x ,")[$a = 2 x ,$]#math.equation(block: false, alt: "b equals −3 y")[$b = −3 y$] and #math.equation(block: false, alt: "n equals 4.")[$n = 4 .$] #math.equation(block: true, alt: "open parenthesis 2 x minus 3 y close parenthesis to the power 4 equals open parenthesis 4; 0 close parenthesis open parenthesis 2 x close parenthesis to the power 4 plus open parenthesis 4; 1 close parenthesis open parenthesis 2 x close parenthesis to the power 4 minus 1 open parenthesis −3 y close parenthesis to the power 1 plus open parenthesis 4; 2 close parenthesis open parenthesis 2 x close parenthesis to the power 4 minus 2 open parenthesis −3 y close parenthesis squared plus open parenthesis 4; 3 close parenthesis open parenthesis 2 x close parenthesis to the power 4 minus 3 open parenthesis −3 y close parenthesis cubed plus open parenthesis 4; 4 close parenthesis open parenthesis −3 y close parenthesis to the power 4")[$attach(( 2 x − 3 y ), t: 4) = ( 4 \ 0 ) attach(( 2 x ), t: 4) + ( 4 \ 1 ) attach(( 2 x ), t: 4 − 1) attach(( −3 y ), t: 1) + ( 4 \ 2 ) attach(( 2 x ), t: 4 − 2) attach(( −3 y ), t: 2) + ( 4 \ 3 ) attach(( 2 x ), t: 4 − 3) attach(( −3 y ), t: 3) + ( 4 \ 4 ) attach(( −3 y ), t: 4)$]Simplify the exponents. #math.equation(block: true, alt: "open parenthesis 2 x minus 3 y close parenthesis to the power 4 equals open parenthesis 4; 0 close parenthesis open parenthesis 2 x close parenthesis to the power 4 plus open parenthesis 4; 1 close parenthesis open parenthesis 2 x close parenthesis cubed open parenthesis −3 y close parenthesis to the power 1 plus open parenthesis 4; 2 close parenthesis open parenthesis 2 x close parenthesis squared open parenthesis −3 y close parenthesis squared plus open parenthesis 4; 3 close parenthesis open parenthesis 2 x close parenthesis to the power 1 open parenthesis −3 y close parenthesis cubed plus open parenthesis 4; 4 close parenthesis open parenthesis −3 y close parenthesis to the power 4")[$attach(( 2 x − 3 y ), t: 4) = ( 4 \ 0 ) attach(( 2 x ), t: 4) + ( 4 \ 1 ) attach(( 2 x ), t: 3) attach(( −3 y ), t: 1) + ( 4 \ 2 ) attach(( 2 x ), t: 2) attach(( −3 y ), t: 2) + ( 4 \ 3 ) attach(( 2 x ), t: 1) attach(( −3 y ), t: 3) + ( 4 \ 4 ) attach(( −3 y ), t: 4)$]Evaluate the coefficients. Remember, #math.equation(block: false, alt: "open parenthesis n; 1 close parenthesis equals n , open parenthesis n; n close parenthesis equals 1 , open parenthesis n; 0 close parenthesis equals 1.")[$( n \ 1 ) = n , #h(0.5em) ( n \ n ) = 1 , #h(0.5em) ( n \ 0 ) = 1 .$] #math.equation(block: true, alt: "open parenthesis 2 x minus 3 y close parenthesis to the power 4 equals 1 open parenthesis 2 x close parenthesis to the power 4 plus 4 open parenthesis 2 x close parenthesis cubed open parenthesis −3 y close parenthesis to the power 1 plus the fraction 4 ! over 2 ! open parenthesis 2 close parenthesis ! open parenthesis 2 x close parenthesis squared open parenthesis −3 y close parenthesis squared plus the fraction 4 ! over 3 ! open parenthesis 4 minus 3 close parenthesis ! open parenthesis 2 x close parenthesis to the power 1 open parenthesis −3 y close parenthesis cubed plus 1 open parenthesis −3 y close parenthesis to the power 4")[$attach(( 2 x − 3 y ), t: 4) = 1 attach(( 2 x ), t: 4) + 4 attach(( 2 x ), t: 3) attach(( −3 y ), t: 1) + frac(4 !, 2 ! ( 2 ) !) attach(( 2 x ), t: 2) attach(( −3 y ), t: 2) + frac(4 !, 3 ! ( 4 − 3 ) !) attach(( 2 x ), t: 1) attach(( −3 y ), t: 3) + 1 attach(( −3 y ), t: 4)$] #math.equation(block: true, alt: "open parenthesis 2 x minus 3 y close parenthesis to the power 4 equals 16 x to the power 4 plus 4 times 8 x cubed open parenthesis −3 y close parenthesis plus 6 open parenthesis 4 x squared close parenthesis open parenthesis 9 y squared close parenthesis plus 4 open parenthesis 2 x close parenthesis open parenthesis −27 y cubed close parenthesis plus 81 y to the power 4")[$attach(( 2 x − 3 y ), t: 4) = 16 x^(4) + 4 · 8 x^(3) ( −3 y ) + 6 ( 4 x^(2) ) ( 9 y^(2) ) + 4 ( 2 x ) ( −27 y^(3) ) + 81 y^(4)$] #math.equation(block: true, alt: "open parenthesis 2 x minus 3 y close parenthesis to the power 4 equals 16 x to the power 4 minus 96 x cubed y plus 216 x squared y squared minus 216 x y cubed plus 81 y to the power 4")[$attach(( 2 x − 3 y ), t: 4) = 16 x^(4) − 96 x^(3) y + 21 6 x^(2) y^(2) − 216 x y^(3) + 81 y^(4)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis 3 x minus 2 y close parenthesis to the power 5.")[$attach(( 3 x − 2 y ), t: 5) .$] #solutionbox[ #math.equation(block: true, alt: "243 x to the power 5 minus 810 x to the power 4 y plus 1080 x cubed y squared")[$243 x^(5) − 810 x^(4) y + 1080 x^(3) y^(2)$] #linebreak() #math.equation(block: true, alt: "minus 720 x squared y cubed plus 240 x y to the power 4 minus 32 y to the power 5")[$− 720 x^(2) y^(3) + 240 x y^(4) − 32 y^(5)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use the Binomial Theorem to expand #math.equation(block: false, alt: "open parenthesis 4 x minus 3 y close parenthesis to the power 4.")[$attach(( 4 x − 3 y ), t: 4) .$] #solutionbox[ #math.equation(block: true, alt: "256 x to the power 4 minus 768 x cubed y plus 864 x squared y squared")[$256 x^(4) − 768 x^(3) y + 864 x^(2) y^(2)$] #linebreak() #math.equation(block: true, alt: "minus 432 x y cubed plus 81 y to the power 4")[$− 432 x y^(3) + 81 y^(4)$] ] ] The real beauty of the Binomial Theorem is that it gives a formula for any particular term of the expansion without having to compute the whole sum. Let’s look for a pattern in the #strong[Binomial Theorem]. #figure(figph[This figure shows a plus b to the power of n equals n choose 0 times a to the power of n b to the power of 0 plus n choose 1 times a to the power of n minus 1 b to the 1 plus n choose 2 times a to the power of n minus 2 b to the power of 2 plus ellipsis plus n choose r times a to the power of n minus r plus ellipsis plus n choose n times b to the power of n.], alt: "This figure shows a plus b to the power of n equals n choose 0 times a to the power of n b to the power of 0 plus n choose 1 times a to the power of n minus 1 b to the 1 plus n choose 2 times a to the power of n minus 2 b to the power of 2 plus ellipsis plus n choose r times a to the power of n minus r plus ellipsis plus n choose n times b to the power of n.", caption: none) Notice, that in each case the exponent on the #emph[b] is one less than the number of the term. The #math.equation(block: false, alt: "open parenthesis r plus 1 close parenthesis s t")[$( r + 1 ) s t$] term is the term where the exponent of #emph[b] is #emph[r]. So we can use the format of the #math.equation(block: false, alt: "open parenthesis r plus 1 close parenthesis s t")[$( r + 1 ) s t$] term to find the value of a specific term. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Find a Specific Term in a Binomial Expansion] The #math.equation(block: false, alt: "open parenthesis r plus 1 close parenthesis to the power s t")[$attach(( r + 1 ), t: s t)$] term in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$attach(( a + b ), t: n)$] is #math.equation(block: true, alt: "open parenthesis n; r close parenthesis a to the power n minus r b to the power r")[$( n \ r ) a^(n − r) b^(r)$] ] #examplebox("Example 8")[][ Find the fourth term of #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power 7.")[$attach(( x + y ), t: 7) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([In our pattern, #math.equation(block: false, alt: "n equals 7 , a equals x")[$n = 7 , #h(0.2em) a = x$] and #math.equation(block: false, alt: "b equals y .")[$b = y .$]], [#figure(figph[Two binomial expressions are shown: (a + b) in red raised to the power of n, and (x + y) in black raised to the power of 7, illustrating algebraic notation.], alt: "Two binomial expressions are shown: (a + b) in red raised to the power of n, and (x + y) in black raised to the power of 7, illustrating algebraic notation.", caption: none)]), [We are looking for the fourth term. #linebreak() #math.equation(block: false, alt: "Since r plus 1, equals, 4 ,; then r, equals, 3.")[$#h(1em) "Since" #h(0.2em) r + 1 & = & 4 , \ "then" #h(0.2em) r & = & 3 .$]], [], [Write the formula.], [#figure(figph[The image shows a mathematical expression: the binomial coefficient 'n choose r' multiplied by 'a' raised to the power of 'n-r' and 'b' raised to the power of 'r', often seen in binomial expansion.], alt: "The image shows a mathematical expression: the binomial coefficient 'n choose r' multiplied by 'a' raised to the power of 'n-r' and 'b' raised to the power of 'r', often seen in binomial expansion.", caption: none)], [Substitute in the values, #math.equation(block: false, alt: "n equals 7 ,")[$n = 7 ,$]#math.equation(block: false, alt: "r equals 3 ,")[$r = 3 ,$]#math.equation(block: false, alt: "a equals x ,")[$a = x ,$] and #math.equation(block: false, alt: "b equals y .")[$b = y .$]], [#figure(figph[A mathematical expression showing the binomial coefficient (7 choose 3) multiplied by a raised to the power of (7-3) and b cubed, representing a term from a binomial expansion.], alt: "A mathematical expression showing the binomial coefficient (7 choose 3) multiplied by a raised to the power of (7-3) and b cubed, representing a term from a binomial expansion.", caption: none)], [#figure(figph[A mathematical formula stating 'Use (n over r) = n! / (r!(n - r)!)', which defines the combination formula (n choose r).], alt: "A mathematical formula stating 'Use (n over r) = n! / (r!(n - r)!)', which defines the combination formula (n choose r).", caption: none)], [#figure(figph[A mathematical expression featuring factorials, exponents, and variables is displayed against a white background. The expression reads: '7! / (3!4!) a^(7-3) b^3'.], alt: "A mathematical expression featuring factorials, exponents, and variables is displayed against a white background. The expression reads: '7! / (3!4!) a^(7-3) b^3'.", caption: none)], [Simplify.], [#figure(figph[A mathematical expression: (7\*6\*5\*4!)/(4!\*3\*2\*1) \* a^4b^3. Cancellation of 4! and (3\*2\*1) with 6 simplifies the fraction to 7\*5, resulting in 35a^4b^3.], alt: "A mathematical expression: (7*6*5*4!)/(4!*3*2*1) * a^4b^3. Cancellation of 4! and (3*2*1) with 6 simplifies the fraction to 7*5, resulting in 35a^4b^3.", caption: none)], [Simplify.], [#figure(figph[The image displays the algebraic expression 35a^4b^3 on a white background, representing a monomial with a coefficient and two variables raised to positive integer exponents.], alt: "The image displays the algebraic expression 35a^4b^3 on a white background, representing a monomial with a coefficient and two variables raised to positive integer exponents.", caption: none)], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the third term of #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power 6.")[$attach(( x + y ), t: 6) .$] #solutionbox[ #math.equation(block: true, alt: "15 x to the power 4 y squared")[$15 x^(4) y^(2)$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the fifth term of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 8.")[$attach(( a + b ), t: 8) .$] #solutionbox[ #math.equation(block: true, alt: "70 a to the power 4 b to the power 4")[$70 a^(4) b^(4)$] ] ] #examplebox("Example 9")[][ Find the coefficient of the #math.equation(block: false, alt: "x to the power 6")[$x^(6)$] term of #math.equation(block: false, alt: "open parenthesis x plus 3 close parenthesis to the power 9.")[$attach(( x + 3 ), t: 9) .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([In our pattern, then #math.equation(block: false, alt: "n equals 9 ,")[$n = 9 ,$]#math.equation(block: false, alt: "a equals x ,")[$a = x ,$] and #math.equation(block: false, alt: "b equals 3.")[$b = 3 .$]], [#figure(figph[The image displays two binomial expressions: the general form (a+b)^n in red, and a specific example (x+3)^9 in black, illustrating the concept of binomial expansion.], alt: "The image displays two binomial expressions: the general form (a+b)^n in red, and a specific example (x+3)^9 in black, illustrating the concept of binomial expansion.", caption: none)]), [We are looking for the coefficient of the #math.equation(block: false, alt: "x to the power 6")[$x^(6)$] term. #linebreak()     Since #math.equation(block: false, alt: "a equals x ,")[$a = x ,$] and #math.equation(block: false, alt: "x to the power 9 minus r equals x to the power 6 ,")[$x^(9 − r) = x^(6) ,$] we know #math.equation(block: false, alt: "r equals 3.")[$r = 3 .$]], [], [Write the formula.], [#figure(figph[A mathematical expression featuring the binomial coefficient, represented as (n choose r), multiplied by a raised to the power of (n-r) and b raised to the power of r. This is a common term from binomial expansion.], alt: "A mathematical expression featuring the binomial coefficient, represented as (n choose r), multiplied by a raised to the power of (n-r) and b raised to the power of r. This is a common term from binomial expansion.", caption: none)], [Substitute in the values, #math.equation(block: false, alt: "n equals 9 ,")[$n = 9 ,$]#math.equation(block: false, alt: "r equals 3 ,")[$r = 3 ,$]#math.equation(block: false, alt: "a equals x ,")[$a = x ,$] and #math.equation(block: false, alt: "b equals 3.")[$b = 3 .$]], [#figure(figph[A mathematical expression featuring a binomial coefficient (9 choose 3), variable x raised to the power of (9-3), and the number 3 raised to the power of 3, all multiplied together.], alt: "A mathematical expression featuring a binomial coefficient (9 choose 3), variable x raised to the power of (9-3), and the number 3 raised to the power of 3, all multiplied together.", caption: none)], [#figure(figph[The image displays the formula for combinations, written as 'Use (n over r) = n! / (r!(n-r)!)', which defines how to calculate the number of ways to choose r items from a set of n items without regard to the order of selection.], alt: "The image displays the formula for combinations, written as 'Use (n over r) = n! / (r!(n-r)!)', which defines how to calculate the number of ways to choose r items from a set of n items without regard to the order of selection.", caption: none)], [#figure(figph[A mathematical expression featuring a fraction with factorials (9!/(3!6!)), multiplied by X raised to the power of (theta-3), and then by 3 raised to the power of 3.], alt: "A mathematical expression featuring a fraction with factorials (9!/(3!6!)), multiplied by X raised to the power of (theta-3), and then by 3 raised to the power of 3.", caption: none)], [Simplify.], [#figure(figph[The mathematical expression (9 \* 8 \* 7 \* 6!) / (3! \* 6!) multiplied by x^6 and 27, demonstrating algebraic and factorial notation.], alt: "The mathematical expression (9 * 8 * 7 * 6!) / (3! * 6!) multiplied by x^6 and 27, demonstrating algebraic and factorial notation.", caption: none)], [Simplify.], [#figure(figph[A mathematical expression featuring the numbers 84 and 27, multiplied by 'x' raised to the power of 6, on a white background.], alt: "A mathematical expression featuring the numbers 84 and 27, multiplied by 'x' raised to the power of 6, on a white background.", caption: none)], [Simplify.], [#figure(figph[The mathematical expression 2268x^6 is displayed in a dark gray font against a plain white background, appearing as part of a larger equation or problem.], alt: "The mathematical expression 2268x^6 is displayed in a dark gray font against a plain white background, appearing as part of a larger equation or problem.", caption: none)], [], [The coefficient of the #math.equation(block: false, alt: "x to the power 6")[$x^(6)$] term is 2268.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the coefficient of the #math.equation(block: false, alt: "x to the power 5")[$x^(5)$] term of #math.equation(block: false, alt: "open parenthesis x plus 4 close parenthesis to the power 8.")[$attach(( x + 4 ), t: 8) .$] #solutionbox[ 3,584 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the coefficient of the #math.equation(block: false, alt: "x to the power 4")[$x^(4)$] term of #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis to the power 7.")[$attach(( x + 2 ), t: 7) .$] #solutionbox[ 280 ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access these online resources for additional instruction and practice with sequences. - #link("https://openstax.org/l/37binexpanpastr")[Binomial Expansion Using Pascal’s Triangle] - #link("https://openstax.org/l/37binocoeffic")[Binomial Coefficients] ] === Key Concepts - #strong[Patterns in the expansion of] #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$attach(( a + b ), t: n)$] - The number of terms is #math.equation(block: false, alt: "n plus 1.")[$n + 1 .$] - The first term is #math.equation(block: false, alt: "a to the power n")[$a^(n)$] and the last term is #math.equation(block: false, alt: "b to the power n .")[$b^(n) .$] - The exponents on #emph[a] decrease by one on each term going left to right. - The exponents on #emph[b] increase by one on each term going left to right. - The sum of the exponents on any term is #emph[n]. - #strong[Pascal’s Triangle] #figure(figph[This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1], alt: "This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1", caption: none) - #strong[Binomial Coefficient] #math.equation(block: false, alt: "open parenthesis n; r close parenthesis")[$( n \ r )$] #strong[:] A binomial coefficient #math.equation(block: false, alt: "open parenthesis n; r close parenthesis ,")[$( n \ r ) ,$] where #emph[r] and #emph[n] are integers with #math.equation(block: false, alt: "0 less than or equal to r less than or equal to n ,")[$0 ≤ r ≤ n ,$] is defined as #linebreak() #math.equation(block: true, alt: "open parenthesis n; r close parenthesis equals the fraction n ! over r ! open parenthesis n minus r close parenthesis !")[$( n \ r ) = frac(n !, r ! ( n − r ) !)$] #linebreak() We read #math.equation(block: false, alt: "open parenthesis n; r close parenthesis")[$( n \ r )$] as “#emph[n] choose #emph[r]” or “#emph[n] taken #emph[r] at a time”. - #strong[Properties of Binomial Coefficients] #linebreak() #math.equation(block: false, alt: "open parenthesis n; 1 close parenthesis equals n, open parenthesis n; n close parenthesis equals 1, open parenthesis n; 0 close parenthesis equals 1")[$( n \ 1 ) = n & & & & ( n \ n ) = 1 & & & & ( n \ 0 ) = 1$] - #strong[Binomial Theorem:] For any real numbers #emph[a], #emph[b], and positive integer #emph[n], #linebreak() #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals open parenthesis n; 0 close parenthesis a to the power n plus open parenthesis n; 1 close parenthesis a to the power n minus 1 b to the power 1 plus open parenthesis n; 2 close parenthesis a to the power n minus 2 b squared plus ... plus open parenthesis n; r close parenthesis a to the power n minus r b to the power r plus ... plus open parenthesis n; n close parenthesis b to the power n")[$attach(( a + b ), t: n) = ( n \ 0 ) a^(n) + ( n \ 1 ) a^(n − 1) b^(1) + ( n \ 2 ) a^(n − 2) b^(2) + ... + ( n \ r ) a^(n − r) b^(r) + ... + ( n \ n ) b^(n)$] === Section Exercises ==== Practice Makes Perfect #strong[Use Pascal’s Triangle to Expand a Binomial] In the following exercises, expand each binomial using Pascal’s Triangle. #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis to the power 4")[$attach(( x + y ), t: 4)$] #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 8")[$attach(( a + b ), t: 8)$] #solutionbox[ #math.equation(block: true, alt: "a to the power 8 plus 8 a to the power 7 b plus 28 a to the power 6 b squared plus 56 a to the power 5 b cubed")[$a^(8) + 8 a^(7) b + 28 a^(6) b^(2) + 56 a^(5) b^(3)$] #linebreak() #math.equation(block: true, alt: "plus 70 a to the power 4 b to the power 4 plus 56 a cubed b to the power 5 plus 28 a squared b to the power 6")[$+ 70 a^(4) b^(4) + 56 a^(3) b^(5) + 28 a^(2) b^(6)$] #linebreak() #math.equation(block: true, alt: "plus 8 a b to the power 7 plus b to the power 8")[$+ 8 a b^(7) + b^(8)$] ] #math.equation(block: true, alt: "open parenthesis m plus n close parenthesis to the power 10")[$attach(( m + n ), t: 10)$] #math.equation(block: true, alt: "open parenthesis p plus q close parenthesis to the power 9")[$attach(( p + q ), t: 9)$] #solutionbox[ #math.equation(block: true, alt: "p to the power 9 plus 9 p to the power 8 q plus 36 p to the power 7 q squared plus 84 p to the power 6 q cubed")[$p^(9) + 9 p^(8) q + 36 p^(7) q^(2) + 84 p^(6) q^(3)$] #linebreak() #math.equation(block: true, alt: "plus 126 p to the power 5 q to the power 4 plus 126 p to the power 4 q to the power 5 plus 84 p cubed q to the power 6")[$+ 126 p^(5) q^(4) + 126 p^(4) q^(5) + 84 p^(3) q^(6)$] #linebreak() #math.equation(block: true, alt: "plus 36 p squared q to the power 7 plus 9 p q to the power 8 plus q to the power 9")[$+ 36 p^(2) q^(7) + 9 p q^(8) + q^(9)$] ] #math.equation(block: true, alt: "open parenthesis x minus y close parenthesis to the power 5")[$attach(( x − y ), t: 5)$] #math.equation(block: true, alt: "open parenthesis a minus b close parenthesis to the power 6")[$attach(( a − b ), t: 6)$] #solutionbox[ #math.equation(block: true, alt: "a to the power 6 minus 6 a to the power 5 b plus 15 a to the power 4 b squared minus 20 a cubed b cubed")[$a^(6) − 6 a^(5) b + 15 a^(4) b^(2) − 20 a^(3) b^(3)$] #linebreak() #math.equation(block: true, alt: "plus 15 a squared b to the power 4 minus 6 a b to the power 5 plus b to the power 6")[$+ 15 a^(2) b^(4) − 6 a b^(5) + b^(6)$] ] #math.equation(block: true, alt: "open parenthesis x plus 4 close parenthesis to the power 4")[$attach(( x + 4 ), t: 4)$] #math.equation(block: true, alt: "open parenthesis x plus 5 close parenthesis cubed")[$attach(( x + 5 ), t: 3)$] #solutionbox[ #math.equation(block: true, alt: "x cubed plus 15 x squared plus 75 x plus 125")[$x^(3) + 15 x^(2) + 75 x + 125$] ] #math.equation(block: true, alt: "open parenthesis y plus 2 close parenthesis to the power 5")[$attach(( y + 2 ), t: 5)$] #math.equation(block: true, alt: "open parenthesis y plus 1 close parenthesis to the power 7")[$attach(( y + 1 ), t: 7)$] #solutionbox[ #math.equation(block: true, alt: "y to the power 7 plus 7 y to the power 6 plus 21 y to the power 5 plus 35 y to the power 4 plus 35 y cubed")[$y^(7) + 7 y^(6) + 21 y^(5) + 35 y^(4) + 35 y^(3)$] #linebreak() #math.equation(block: true, alt: "plus 21 y squared plus 7 y plus 1")[$+ 21 y^(2) + 7 y + 1$] ] #math.equation(block: true, alt: "open parenthesis z minus 3 close parenthesis to the power 5")[$attach(( z − 3 ), t: 5)$] #math.equation(block: true, alt: "open parenthesis z minus 2 close parenthesis to the power 6")[$attach(( z − 2 ), t: 6)$] #solutionbox[ #math.equation(block: true, alt: "z to the power 6 minus 12 z to the power 5 plus 60 z to the power 4 minus 160 z cubed plus 240 z squared")[$z^(6) − 12 z^(5) + 60 z^(4) − 160 z^(3) + 240 z^(2)$] #linebreak() #math.equation(block: true, alt: "minus 192 z plus 64")[$− 192 z + 64$] ] #math.equation(block: true, alt: "open parenthesis 4 x minus 1 close parenthesis cubed")[$attach(( 4 x − 1 ), t: 3)$] #math.equation(block: true, alt: "open parenthesis 3 x minus 1 close parenthesis to the power 5")[$attach(( 3 x − 1 ), t: 5)$] #solutionbox[ #math.equation(block: true, alt: "243 x to the power 5 minus 405 x to the power 4 plus 270 x cubed minus 90 x squared")[$243 x^(5) − 405 x^(4) + 270 x^(3) − 90 x^(2)$] #linebreak() #math.equation(block: true, alt: "plus 15 x minus 1")[$+ 15 x − 1$] ] #math.equation(block: true, alt: "open parenthesis 3 x minus 4 close parenthesis to the power 4")[$attach(( 3 x − 4 ), t: 4)$] #math.equation(block: true, alt: "open parenthesis 3 x minus 5 close parenthesis cubed")[$attach(( 3 x − 5 ), t: 3)$] #solutionbox[ #math.equation(block: true, alt: "27 x cubed minus 135 x squared plus 225 x minus 125")[$27 x^(3) − 135 x^(2) + 225 x − 125$] ] #math.equation(block: true, alt: "open parenthesis 2 x plus 3 y close parenthesis cubed")[$attach(( 2 x + 3 y ), t: 3)$] #math.equation(block: true, alt: "open parenthesis 3 x plus 5 y close parenthesis cubed")[$attach(( 3 x + 5 y ), t: 3)$] #solutionbox[ #math.equation(block: true, alt: "27 x cubed plus 135 x squared y plus 225 x y squared plus 125 y cubed")[$27 x^(3) + 135 x^(2) y + 225 x y^(2) + 125 y^(3)$] ] #strong[Evaluate a Binomial Coefficient] In the following exercises, evaluate. ⓐ #math.equation(block: false, alt: "open parenthesis 8; 1 close parenthesis")[$( 8 \ 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 10; 10 close parenthesis")[$( 10 \ 10 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 6; 0 close parenthesis")[$( 6 \ 0 )$] ⓓ #math.equation(block: false, alt: "open parenthesis 9; 3 close parenthesis")[$( 9 \ 3 )$] ⓐ #math.equation(block: false, alt: "open parenthesis 7; 1 close parenthesis")[$( 7 \ 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 4; 4 close parenthesis")[$( 4 \ 4 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 3; 0 close parenthesis")[$( 3 \ 0 )$] ⓓ #math.equation(block: false, alt: "open parenthesis 10; 8 close parenthesis")[$( 10 \ 8 )$] #solutionbox[ ⓐ 7 ⓑ 1 ⓒ 1 ⓓ 45 ] ⓐ #math.equation(block: false, alt: "open parenthesis 3; 1 close parenthesis")[$( 3 \ 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 9; 9 close parenthesis")[$( 9 \ 9 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 7; 0 close parenthesis")[$( 7 \ 0 )$] ⓓ #math.equation(block: false, alt: "open parenthesis 5; 3 close parenthesis")[$( 5 \ 3 )$] ⓐ #math.equation(block: false, alt: "open parenthesis 4; 1 close parenthesis")[$( 4 \ 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis 5; 5 close parenthesis")[$( 5 \ 5 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 8; 0 close parenthesis")[$( 8 \ 0 )$] ⓓ #math.equation(block: false, alt: "open parenthesis 11; 9 close parenthesis")[$( 11 \ 9 )$] #solutionbox[ ⓐ 4 ⓑ 1 ⓒ 1 ⓓ 55 ] #strong[Use the Binomial Theorem to Expand a Binomial] In the following exercises, expand each binomial. #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis cubed")[$attach(( x + y ), t: 3)$] #math.equation(block: true, alt: "open parenthesis m plus n close parenthesis to the power 5")[$attach(( m + n ), t: 5)$] #solutionbox[ #math.equation(block: true, alt: "m to the power 5 plus 5 m to the power 4 n plus 10 m cubed n squared plus 10 m squared n cubed")[$m^(5) + 5 m^(4) n + 10 m^(3) n^(2) + 10 m^(2) n^(3)$] #linebreak() #math.equation(block: true, alt: "plus 5 m n to the power 4 plus n to the power 5")[$+ 5 m n^(4) + n^(5)$] ] #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 6")[$attach(( a + b ), t: 6)$] #math.equation(block: true, alt: "open parenthesis s plus t close parenthesis to the power 7")[$attach(( s + t ), t: 7)$] #solutionbox[ #math.equation(block: true, alt: "s to the power 7 plus 7 s to the power 6 t plus 21 s to the power 5 t squared plus 35 s to the power 4 t cubed")[$s^(7) + 7 s^(6) t + 21 s^(5) t^(2) + 35 s^(4) t^(3)$] #linebreak() #math.equation(block: true, alt: "plus 35 s cubed t to the power 4 plus 21 s squared t to the power 5 plus 7 s t to the power 6 plus t to the power 7")[$+ 35 s^(3) t^(4) + 21 s^(2) t^(5) + 7 s t^(6) + t^(7)$] ] #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis to the power 4")[$attach(( x − 2 ), t: 4)$] #math.equation(block: true, alt: "open parenthesis y minus 3 close parenthesis to the power 4")[$attach(( y − 3 ), t: 4)$] #solutionbox[ #math.equation(block: true, alt: "y to the power 4 minus 12 y cubed plus 54 y squared minus 108 y plus 81")[$y^(4) − 12 y^(3) + 54 y^(2) − 108 y + 81$] ] #math.equation(block: true, alt: "open parenthesis p minus 1 close parenthesis to the power 5")[$attach(( p − 1 ), t: 5)$] #math.equation(block: true, alt: "open parenthesis q minus 4 close parenthesis cubed")[$attach(( q − 4 ), t: 3)$] #solutionbox[ #math.equation(block: true, alt: "q cubed minus 12 q squared plus 48 q minus 64")[$q^(3) − 12 q^(2) + 48 q − 64$] ] #math.equation(block: true, alt: "open parenthesis 3 x minus y close parenthesis to the power 5")[$attach(( 3 x − y ), t: 5)$] #math.equation(block: true, alt: "open parenthesis 5 x minus 2 y close parenthesis to the power 4")[$attach(( 5 x − 2 y ), t: 4)$] #solutionbox[ #math.equation(block: true, alt: "625 x to the power 4 minus 1000 x cubed y plus 600 x squared y squared")[$625 x^(4) − 1000 x^(3) y + 600 x^(2) y^(2)$] #linebreak() #math.equation(block: true, alt: "minus 160 x y cubed plus 16 y to the power 4")[$− 160 x y^(3) + 16 y^(4)$] ] #math.equation(block: true, alt: "open parenthesis 2 x plus 5 y close parenthesis to the power 4")[$attach(( 2 x + 5 y ), t: 4)$] #math.equation(block: true, alt: "open parenthesis 3 x plus 4 y close parenthesis to the power 5")[$attach(( 3 x + 4 y ), t: 5)$] #solutionbox[ #math.equation(block: true, alt: "243 x to the power 5 plus 1620 x to the power 4 y plus 4320 x cubed y squared")[$243 x^(5) + 1620 x^(4) y + 4320 x^(3) y^(2)$] #linebreak() #math.equation(block: true, alt: "plus 5760 x squared y cubed plus 3840 x y to the power 4 plus 1024 y to the power 5")[$+ 5760 x^(2) y^(3) + 3840 x y^(4) + 1024 y^(5)$] ] In the following exercises, find the indicated term in the expansion of the binomial. Sixth term of #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power 10")[$attach(( x + y ), t: 10)$] Fifth term of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 9")[$attach(( a + b ), t: 9)$] #solutionbox[ #math.equation(block: true, alt: "126 a to the power 5 b to the power 4")[$126 a^(5) b^(4)$] ] Fourth term of #math.equation(block: false, alt: "open parenthesis x minus y close parenthesis to the power 8")[$attach(( x − y ), t: 8)$] Seventh term of #math.equation(block: false, alt: "open parenthesis x minus y close parenthesis to the power 11")[$attach(( x − y ), t: 11)$] #solutionbox[ #math.equation(block: true, alt: "462 x to the power 5 y to the power 6")[$462 x^(5) y^(6)$] ] In the following exercises, find the coefficient of the indicated term in the expansion of the binomial. #math.equation(block: false, alt: "y cubed")[$y^(3)$] term of #math.equation(block: false, alt: "open parenthesis y plus 5 close parenthesis to the power 4")[$attach(( y + 5 ), t: 4)$] #math.equation(block: false, alt: "x to the power 6")[$x^(6)$] term of #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis to the power 8")[$attach(( x + 2 ), t: 8)$] #solutionbox[ 112 ] #math.equation(block: false, alt: "x to the power 5")[$x^(5)$] term of #math.equation(block: false, alt: "open parenthesis x minus 4 close parenthesis to the power 6")[$attach(( x − 4 ), t: 6)$] #math.equation(block: false, alt: "x to the power 7")[$x^(7)$] term of #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis to the power 9")[$attach(( x − 3 ), t: 9)$] #solutionbox[ 324 ] #math.equation(block: false, alt: "a to the power 4 b squared")[$a^(4) b^(2)$] term of #math.equation(block: false, alt: "open parenthesis 2 a plus b close parenthesis to the power 6")[$attach(( 2 a + b ), t: 6)$] #math.equation(block: false, alt: "p to the power 5 q to the power 4")[$p^(5) q^(4)$] term of #math.equation(block: false, alt: "open parenthesis 3 p plus q close parenthesis to the power 9")[$attach(( 3 p + q ), t: 9)$] #solutionbox[ 30,618 ] ==== Writing Exercises In your own words explain how to find the rows of the Pascal’s Triangle. Write the first five rows of Pascal’s Triangle. In your own words, explain the pattern of exponents for each variable in the expansion of. #solutionbox[ Answers will vary. ] In your own words, explain the difference between #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$attach(( a + b ), t: n)$] and #math.equation(block: false, alt: "open parenthesis a minus b close parenthesis to the power n .")[$attach(( a − b ), t: n) .$] In your own words, explain how to find a specific term in the expansion of a binomial without expanding the whole thing. Use an example to help explain. #solutionbox[ Answers will vary. ] ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This figure shows a table with four rows and four columns. The first row is the header row and reads. “I can”, “Confidently”, “With some help” and “No, I don’t get it”. The first column, beginning at the second row reads, “Use Pascal’s Triangle to Expand a Binomial”, “Evaluate a Binomial Coefficient” and “Use the Binomial Theorem to Expand a Binomial”. The remaining columns are blank.], alt: "This figure shows a table with four rows and four columns. The first row is the header row and reads. “I can”, “Confidently”, “With some help” and “No, I don’t get it”. The first column, beginning at the second row reads, “Use Pascal’s Triangle to Expand a Binomial”, “Evaluate a Binomial Coefficient” and “Use the Binomial Theorem to Expand a Binomial”. The remaining columns are blank.", caption: none) ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this? === Chapter Review Exercises ==== Sequences #strong[Write the First Few Terms of a Sequence] In the following exercises, write the first five terms of the sequence whose general term is given. #math.equation(block: true, alt: "a sub n equals 7 n minus 5")[$a_(n) = 7 n − 5$] #math.equation(block: true, alt: "a sub n equals 3 to the power n plus 4")[$a_(n) = 3^(n) + 4$] #solutionbox[ #math.equation(block: true, alt: "7 , 13 , 31 , 85 , 247")[$7 , 13 , 31 , 85 , 247$] ] #math.equation(block: true, alt: "a sub n equals 2 to the power n plus n")[$a_(n) = 2^(n) + n$] #math.equation(block: true, alt: "a sub n equals the fraction 2 n plus 1 over 4 to the power n")[$a_(n) = frac(2 n + 1, 4^(n))$] #solutionbox[ #math.equation(block: true, alt: "the fraction 3 over 4 , the fraction 5 over 16 , the fraction 7 over 64 , the fraction 9 over 256 , the fraction 11 over 1024")[$frac(3, 4) , frac(5, 16) , frac(7, 64) , frac(9, 256) , frac(11, 1024)$] ] #math.equation(block: true, alt: "a sub n equals the fraction open parenthesis −1 close parenthesis to the power n over n squared")[$a_(n) = frac(attach(( −1 ), t: n), n^(2))$] #strong[Find a Formula for the General Term (#emph[n]th Term) of a Sequence] In the following exercises, find a general term for the sequence whose first five terms are shown. #math.equation(block: true, alt: "9 , 18 , 27 , 36 , 45 , …")[$9 , 18 , 27 , 36 , 45 , "…"$] #solutionbox[ #math.equation(block: true, alt: "a sub n equals 9 n")[$a_(n) = 9 n$] ] #math.equation(block: true, alt: "−5 , −4 , −3 , −2 , −1 , …")[$−5 , −4 , −3 , −2 , −1 , "…"$] #math.equation(block: true, alt: "the fraction 1 over e cubed , the fraction 1 over e squared , the fraction 1 over e , 1 , e , …")[$frac(1, e^(3)) , frac(1, e^(2)) , frac(1, e) , 1 , e , "…"$] #solutionbox[ #math.equation(block: true, alt: "a sub n equals e to the power n minus 4")[$a_(n) = e^(n − 4)$] ] #math.equation(block: true, alt: "1 , −8 , 27 , −64 , 125 , …")[$1 , −8 , 27 , −64 , 125 , "…"$] #math.equation(block: true, alt: "minus the fraction 1 over 3 , minus the fraction 1 over 2 , minus the fraction 3 over 5 , minus the fraction 2 over 3 , minus the fraction 5 over 7 , …")[$− frac(1, 3) , − frac(1, 2) , − frac(3, 5) , − frac(2, 3) , − frac(5, 7) , "…"$] #solutionbox[ #math.equation(block: true, alt: "a sub n equals minus the fraction n over n plus 2")[$a_(n) = − frac(n, n + 2)$] ] #strong[Use Factorial Notation] In the following exercises, using factorial notation, write the first five terms of the sequence whose general term is given. #math.equation(block: true, alt: "a sub n equals 4 n !")[$a_(n) = 4 n !$] #math.equation(block: true, alt: "a sub n equals the fraction n ! over open parenthesis n plus 2 close parenthesis !")[$a_(n) = frac(n !, ( n + 2 ) "!")$] #solutionbox[ #math.equation(block: true, alt: "the fraction 1 over 6 , the fraction 1 over 12 , the fraction 1 over 20 , the fraction 1 over 30 , the fraction 1 over 42")[$frac(1, 6) , frac(1, 12) , frac(1, 20) , frac(1, 30) , frac(1, 42)$] ] #math.equation(block: true, alt: "a sub n equals the fraction open parenthesis n minus 1 close parenthesis ! over open parenthesis n plus 1 close parenthesis squared")[$a_(n) = frac(( n − 1 ) !, attach(( n + 1 ), t: "2"))$] #strong[Find the Partial Sum] In the following exercises, expand the partial sum and find its value. #math.equation(block: true, alt: "∑ i equals 17 open parenthesis 2 i minus 5 close parenthesis")[$display(limits(∑)^(7)_(i = 1) ( 2 i) − 5 )$] #solutionbox[ #math.equation(block: true, alt: "−3 plus open parenthesis −1 close parenthesis plus 1 plus 3 plus 5")[$−3 + ( −1 ) + 1 + 3 + 5$] #linebreak() #math.equation(block: true, alt: "plus 7 plus 9 equals 21")[$+ 7 + 9 = 21$] ] #math.equation(block: true, alt: "∑ i equals 135 to the power i")[$display(limits(∑)^(3)_(i = 1) 5^(i))$] #math.equation(block: true, alt: "∑ k equals 04 the fraction 4 over k !")[$display(limits(∑)^(4)_(k = 0) frac(4, k !))$] #solutionbox[ #math.equation(block: true, alt: "4 plus 4 plus 2 plus the fraction 2 over 3 plus the fraction 1 over 6 equals the fraction 65 over 6")[$4 + 4 + 2 + frac(2, 3) + frac(1, 6) = frac(65, 6)$] ] #math.equation(block: true, alt: "∑ k equals 14 open parenthesis k plus 1 close parenthesis open parenthesis 2 k plus 1 close parenthesis")[$display(limits(∑)^(4)_(k = 1) ( k + 1 ) ( 2 k + 1 ))$] #strong[Use Summation Notation to write a Sum] In the following exercises, write each sum using summation notation. #math.equation(block: true, alt: "minus the fraction 1 over 3 plus the fraction 1 over 9 minus the fraction 1 over 27 plus the fraction 1 over 81 minus the fraction 1 over 243")[$− frac(1, 3) + frac(1, 9) − frac(1, 27) + frac(1, 81) − frac(1, 243)$] #solutionbox[ #math.equation(block: true, alt: "∑ n equals 15 open parenthesis −1 close parenthesis to the power n the fraction 1 over 3 to the power n")[$display(limits(∑)^(5)_(n = 1) attach(( −1 ), t: n) frac(1, 3^(n)))$] ] #math.equation(block: true, alt: "4 minus 8 plus 12 minus 16 plus 20 minus 24")[$4 − 8 + 12 − 16 + 20 − 24$] #math.equation(block: true, alt: "4 plus 2 plus the fraction 4 over 3 plus 1 plus the fraction 4 over 5")[$4 + 2 + frac(4, 3) + 1 + frac(4, 5)$] #solutionbox[ #math.equation(block: true, alt: "∑ n equals 15 the fraction 4 over n")[$display(limits(∑)^(5)_(n = 1) frac(4, n))$] ] ==== Arithmetic Sequences #strong[Determine if a Sequence is Arithmetic] In the following exercises, determine if each sequence is arithmetic, and if so, indicate the common difference. #math.equation(block: true, alt: "1 , 2 , 4 , 8 , 16 , 32 , …")[$1 , 2 , 4 , 8 , 16 , 32 , "…"$] #math.equation(block: true, alt: "−7 , −1 , 5 , 11 , 17 , 23 , …")[$−7 , −1 , 5 , 11 , 17 , 23 , "…"$] #solutionbox[ The sequence is arithmetic with common difference #math.equation(block: false, alt: "d equals 6.")[$d = 6 .$] ] #math.equation(block: true, alt: "13 , 9 , 5 , 1 , −3 , −7 , …")[$13 , 9 , 5 , 1 , −3 , −7 , "…"$] In the following exercises, write the first five terms of each arithmetic sequence with the given first term and common difference. #math.equation(block: false, alt: "a sub 1 equals 5")[$a_(1) = 5$] and #math.equation(block: false, alt: "d equals 3")[$d = 3$] #solutionbox[ #math.equation(block: true, alt: "5 , 8 , 11 , 14 , 17")[$5 , 8 , 11 , 14 , 17$] ] #math.equation(block: false, alt: "a sub 1 equals 8")[$a_(1) = 8$] and #math.equation(block: false, alt: "d equals −2")[$d = −2$] #math.equation(block: false, alt: "a sub 1 equals −13")[$a_(1) = −13$] and #math.equation(block: false, alt: "d equals 6")[$d = 6$] #solutionbox[ #math.equation(block: true, alt: "−13 , −7 , −1 , 5 , 11")[$−13 , −7 , −1 , 5 , 11$] ] #strong[Find the General Term (#emph[n]th Term) of an Arithmetic Sequence] In the following exercises, find the term described using the information provided. Find the twenty-fifth term of a sequence where the first term is five and the common difference is three. Find the thirtieth term of a sequence where the first term is 16 and the common difference is #math.equation(block: false, alt: "−5.")[$−5 .$] #solutionbox[ #math.equation(block: true, alt: "−129")[$−129$] ] Find the seventeenth term of a sequence where the first term is #math.equation(block: false, alt: "−21")[$−21$] and the common difference is two. In the following exercises, find the indicated term and give the formula for the general term. Find the eighteenth term of a sequence where the fifth term is #math.equation(block: false, alt: "12")[$12$] and the common difference is seven. #solutionbox[ #math.equation(block: false, alt: "a sub 18 equals 103.")[$a_(18) = 103 .$] The general term is #math.equation(block: false, alt: "a sub n equals 7 n minus 23.")[$a_(n) = 7 n − 23 .$] ] Find the twenty-first term of a sequence where the seventh term is #math.equation(block: false, alt: "14")[$14$] and the common difference is #math.equation(block: false, alt: "−3.")[$−3 .$] In the following exercises, find the first term and common difference of the sequence with the given terms. Give the formula for the general term. The fifth term is 17 and the fourteenth term is 53. #solutionbox[ #math.equation(block: false, alt: "a sub 1 equals 1 ,")[$a_(1) = 1 ,$]#math.equation(block: false, alt: "d equals 4.")[$d = 4 .$] The general term is #math.equation(block: false, alt: "a sub n equals 4 n minus 3.")[$a_(n) = 4 n − 3 .$] ] The third term is #math.equation(block: false, alt: "−26")[$−26$] and the sixteenth term is #math.equation(block: false, alt: "−91.")[$−91 .$] #strong[Find the Sum of the First #emph[n] Terms of an Arithmetic Sequence] In the following exercises, find the sum of the first 30 terms of each arithmetic sequence. #math.equation(block: true, alt: "7 , 4 , 1 , −2 , −5 , …")[$7 , 4 , 1 , −2 , −5 , "…"$] #solutionbox[ #math.equation(block: true, alt: "−1,095")[$−1,095$] ] #math.equation(block: true, alt: "1 , 6 , 11 , 16 , 21 , …")[$1 , 6 , 11 , 16 , 21 , "…"$] In the following exercises, find the sum of the first fifteen terms of the arithmetic sequence whose general term is given. #math.equation(block: true, alt: "a sub n equals 4 n plus 7")[$a_(n) = 4 n + 7$] #solutionbox[ #math.equation(block: true, alt: "585")[$585$] ] #math.equation(block: true, alt: "a sub n equals −2 n plus 19")[$a_(n) = −2 n + 19$] In the following exercises, find each sum. #math.equation(block: true, alt: "∑ i equals 150 open parenthesis 4 i minus 5 close parenthesis")[$display(limits(∑)^(50)_(i = 1) ( 4 i − 5) )$] #solutionbox[ #math.equation(block: true, alt: "4,850")[$4,850$] ] #math.equation(block: true, alt: "∑ i equals 130 open parenthesis −3 i minus 7 close parenthesis")[$display(limits(∑)^(30)_(i = 1) ( −3 i − 7) )$] #math.equation(block: true, alt: "∑ i equals 135 open parenthesis i plus 10 close parenthesis")[$display(limits(∑)^(35)_(i = 1) ( i + 10) )$] #solutionbox[ #math.equation(block: true, alt: "980")[$980$] ] ==== Geometric Sequences and Series #strong[Determine if a Sequence is Geometric] In the following exercises, determine if the sequence is geometric, and if so, indicate the common ratio. #math.equation(block: true, alt: "3 , 12 , 48 , 192 , 768 , 3072 , …")[$3 , 12 , 48 , 192 , 768 , 3072 , "…"$] #math.equation(block: true, alt: "5 , 10 , 15 , 20 , 25 , 30 , …")[$5 , 10 , 15 , 20 , 25 , 30 , "…"$] #solutionbox[ The sequence is not geometric. ] #math.equation(block: true, alt: "112 , 56 , 28 , 14 , 7 , the fraction 7 over 2 , …")[$112 , 56 , 28 , 14 , 7 , frac(7, 2) , "…"$] #math.equation(block: true, alt: "9 , −18 , 36 , −72 , 144 , −288 , …")[$9 , −18 , 36 , −72 , 144 , −288 , "…"$] #solutionbox[ The sequence is geometric with common ratio #math.equation(block: false, alt: "r equals −2.")[$r = −2 .$] ] In the following exercises, write the first five terms of each geometric sequence with the given first term and common ratio. #math.equation(block: false, alt: "a sub 1 equals −3")[$a_(1) = −3$] and #math.equation(block: false, alt: "r equals 5")[$r = 5$] #math.equation(block: false, alt: "a sub 1 equals 128")[$a_(1) = 128$] and #math.equation(block: false, alt: "r equals the fraction 1 over 4")[$r = frac(1, 4)$] #solutionbox[ #math.equation(block: true, alt: "128 , 32 , 8 , 2 , the fraction 1 over 2")[$128 , 32 , 8 , 2 , frac(1, 2)$] ] #math.equation(block: false, alt: "a sub 1 equals 5")[$a_(1) = 5$] and #math.equation(block: false, alt: "r equals −3")[$r = −3$] #strong[Find the General Term (#emph[n]th Term) of a Geometric Sequence] In the following exercises, find the indicated term of a sequence where the first term and the common ratio is given. Find #math.equation(block: false, alt: "a sub 9")[$a_(9)$] given #math.equation(block: false, alt: "a sub 1 equals 6")[$a_(1) = 6$] and #math.equation(block: false, alt: "r equals 2.")[$r = 2 .$] #solutionbox[ #math.equation(block: true, alt: "1,536")[$1,536$] ] Find #math.equation(block: false, alt: "a sub 11")[$a_(11)$] given #math.equation(block: false, alt: "a sub 1 equals 10,000,000")[$a_(1) = 10,000,000$] and #math.equation(block: false, alt: "r equals 0.1.")[$r = 0.1 .$] In the following exercises, find the indicated term of the given sequence. Find the general term of the sequence. Find #math.equation(block: false, alt: "a sub 12")[$a_(12)$] of the sequence, #math.equation(block: false, alt: "6 , −24 , 96 , −384 , 1536 , −6144 , …")[$6 , −24 , 96 , −384 , 1536 , −6144 , "…"$] #solutionbox[ #math.equation(block: false, alt: "a sub 12 equals −25 , 165 , 824.")[$a_(12) = −25 , 165 , 824 .$] The general term is #math.equation(block: false, alt: "a sub n equals 6 open parenthesis −4 close parenthesis to the power n minus 1.")[$a_(n) = 6 attach(( −4 ), t: n − 1) .$] ] Find #math.equation(block: false, alt: "a sub 9")[$a_(9)$] of the sequence, #math.equation(block: false, alt: "4374 , 1458 , 486 , 162 , 54 , 18 , …")[$4374 , 1458 , 486 , 162 , 54 , 18 , "…"$] #strong[Find the Sum of the First #emph[n] terms of a Geometric Sequence] In the following exercises, find the sum of the first fifteen terms of each geometric sequence. #math.equation(block: true, alt: "−4 , 8 , −16 , 32 , −64 , 128 …")[$−4 , 8 , −16 , 32 , −64 , 128 "…"$] #solutionbox[ −43,692 ] #math.equation(block: true, alt: "3 , 12 , 48 , 192 , 768 , 3072 …")[$3 , 12 , 48 , 192 , 768 , 3072 "…"$] #math.equation(block: true, alt: "3125 , 625 , 125 , 25 , 5 , 1 …")[$3125 , 625 , 125 , 25 , 5 , 1 "…"$] #solutionbox[ #math.equation(block: true, alt: "3906.25")[$3906.25$] ] In the following exercises, find the sum #math.equation(block: true, alt: "∑ i equals 187 open parenthesis 3 close parenthesis to the power i")[$display(limits(∑)^(8)_(i = 1) 7 attach(( 3 ), t: i))$] #math.equation(block: true, alt: "∑ i equals 1624 open parenthesis the fraction 1 over 2 close parenthesis to the power i")[$display(limits(∑)^(6)_(i = 1) 24 attach(( frac(1, 2) ), t: i))$] #solutionbox[ #math.equation(block: true, alt: "the fraction 189 over 8 equals 23.625")[$frac(189, 8) = 23.625$] ] #strong[Find the Sum of an Infinite Geometric Series] In the following exercises, find the sum of each infinite geometric series. #math.equation(block: true, alt: "1 minus the fraction 1 over 3 plus the fraction 1 over 9 minus the fraction 1 over 27 plus the fraction 1 over 81 minus the fraction 1 over 243 plus the fraction 1 over 729 minus …")[$1 − frac(1, 3) + frac(1, 9) − frac(1, 27) + frac(1, 81) − frac(1, 243) + frac(1, 729) − "…"$] #math.equation(block: true, alt: "49 plus 7 plus 1 plus the fraction 1 over 7 plus the fraction 1 over 49 plus the fraction 1 over 343 plus …")[$49 + 7 + 1 + frac(1, 7) + frac(1, 49) + frac(1, 343) + "…"$] #solutionbox[ #math.equation(block: true, alt: "the fraction 343 over 6 approximately equals 57.167")[$frac(343, 6) ≈ 57.167$] ] In the following exercises, write each repeating decimal as a fraction. #math.equation(block: true, alt: "0.8 bar")[$0. accent(8, −)$] #math.equation(block: true, alt: "0.36 bar")[$0. accent(36, —)$] #solutionbox[ #math.equation(block: true, alt: "the fraction 4 over 11")[$frac(4, 11)$] ] #strong[Apply Geometric Sequences and Series in the Real World] In the following exercises, solve the problem. What is the total effect on the economy of a government tax rebate of #math.equation(block: false, alt: "$ 360")[$"$" 360$] to each household in order to stimulate the economy if each household will spend #math.equation(block: false, alt: "60 %")[$60 %$] of the rebate in goods and services? Adam just got his first full-time job after graduating from high school at age 17. He decided to invest #math.equation(block: false, alt: "$ 300")[$"$" 300$] per month in an IRA (an annuity). The interest on the annuity is #math.equation(block: false, alt: "7 %")[$7 %$] which is compounded monthly. How much will be in Adam’s account when he retires at his sixty-seventh birthday? #solutionbox[ #math.equation(block: true, alt: "$ 1,634,421.27")[$"$" 1,634,421.27$] ] ==== Binomial Theorem #strong[Use Pascal’s Triangle to Expand a Binomial] In the following exercises, expand each binomial using Pascal’s Triangle. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 7")[$attach(( a + b ), t: 7)$] #math.equation(block: true, alt: "open parenthesis x minus y close parenthesis to the power 4")[$attach(( x − y ), t: 4)$] #solutionbox[ #math.equation(block: true, alt: "x to the power 4 minus 4 x cubed y plus 6 x squared y squared minus 4 x y cubed plus y to the power 4")[$x^(4) − 4 x^(3) y + 6 x^(2) y^(2) − 4 x y^(3) + y^(4)$] ] #math.equation(block: true, alt: "open parenthesis x plus 6 close parenthesis cubed")[$attach(( x + 6 ), t: 3)$] #math.equation(block: true, alt: "open parenthesis 2 y minus 3 close parenthesis to the power 5")[$attach(( 2 y − 3 ), t: 5)$] #solutionbox[ #math.equation(block: true, alt: "32 y to the power 5 minus 240 y to the power 4 plus 720 y cubed minus 1080 y squared")[$32 y^(5) − 240 y^(4) + 720 y^(3) − 1080 y^(2)$] #linebreak() #math.equation(block: true, alt: "plus 810 y minus 243")[$+ 810 y − 243$] ] #math.equation(block: true, alt: "open parenthesis 7 x plus 2 y close parenthesis cubed")[$attach(( 7 x + 2 y ), t: 3)$] #strong[Evaluate a Binomial Coefficient] In the following exercises, evaluate. ⓐ #math.equation(block: false, alt: "open parenthesis 11; 1 close parenthesis")[$( 11 \ 1 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 12; 12 close parenthesis")[$( 12 \ 12 )$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis 13; 0 close parenthesis")[$( 13 \ 0 )$] #linebreak() ⓓ #math.equation(block: false, alt: "open parenthesis 8; 3 close parenthesis")[$( 8 \ 3 )$] #solutionbox[ ⓐ 11 ⓑ 1 ⓒ 1 ⓓ 56 ] ⓐ #math.equation(block: false, alt: "open parenthesis 7; 1 close parenthesis")[$( 7 \ 1 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 5; 5 close parenthesis")[$( 5 \ 5 )$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis 9; 0 close parenthesis")[$( 9 \ 0 )$] #linebreak() ⓓ #math.equation(block: false, alt: "open parenthesis 9; 5 close parenthesis")[$( 9 \ 5 )$] ⓐ #math.equation(block: false, alt: "open parenthesis 1; 1 close parenthesis")[$( 1 \ 1 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 15; 15 close parenthesis")[$( 15 \ 15 )$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis 4; 0 close parenthesis")[$( 4 \ 0 )$] #linebreak() ⓓ #math.equation(block: false, alt: "open parenthesis 11; 2 close parenthesis")[$( 11 \ 2 )$] #solutionbox[ ⓐ 1 ⓑ 1 ⓒ 1 ⓓ 55 ] #strong[Use the Binomial Theorem to Expand a Binomial] In the following exercises, expand each binomial, using the Binomial Theorem. #math.equation(block: true, alt: "open parenthesis p plus q close parenthesis to the power 6")[$attach(( p + q ), t: 6)$] #math.equation(block: true, alt: "open parenthesis t minus 1 close parenthesis to the power 9")[$attach(( t − 1 ), t: 9)$] #solutionbox[ #math.equation(block: true, alt: "t to the power 9 minus 9 t to the power 8 plus 36 t to the power 7 minus 84 t to the power 6 plus 126 t to the power 5")[$t^(9) − 9 t^(8) + 36 t^(7) − 84 t^(6) + 126 t^(5)$] #linebreak() #math.equation(block: true, alt: "minus 126 t to the power 4 plus 84 t cubed minus 36 t squared plus 9 t minus 1")[$− 126 t^(4) + 84 t^(3) − 36 t^(2) + 9 t − 1$] ] #math.equation(block: true, alt: "open parenthesis 2 x plus 1 close parenthesis to the power 4")[$attach(( 2 x + 1 ), t: 4)$] #math.equation(block: true, alt: "open parenthesis 4 x plus 3 y close parenthesis to the power 4")[$attach(( 4 x + 3 y ), t: 4)$] #solutionbox[ #math.equation(block: true, alt: "256 x to the power 4 plus 768 x cubed y plus 864 x squared y squared")[$256 x^(4) + 768 x^(3) y + 864 x^(2) y^(2)$] #linebreak() #math.equation(block: true, alt: "plus 432 x y cubed plus 81 y to the power 4")[$+ 432 x y^(3) + 81 y^(4)$] ] #math.equation(block: true, alt: "open parenthesis x minus 3 y close parenthesis to the power 5")[$attach(( x − 3 y ), t: 5)$] In the following exercises, find the indicated term in the expansion of the binomial. Seventh term of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 9")[$attach(( a + b ), t: 9)$] #solutionbox[ #math.equation(block: true, alt: "84 a cubed b to the power 6")[$84 a^(3) b^(6)$] ] Third term of #math.equation(block: false, alt: "open parenthesis x minus y close parenthesis to the power 7")[$attach(( x − y ), t: 7)$] In the following exercises, find the coefficient of the indicated term in the expansion of the binomial. #math.equation(block: false, alt: "y to the power 4")[$y^(4)$] term of #math.equation(block: false, alt: "open parenthesis y plus 3 close parenthesis to the power 6")[$attach(( y + 3 ), t: 6)$] #solutionbox[ 135 ] #math.equation(block: false, alt: "x to the power 5")[$x^(5)$] term of #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis to the power 8")[$attach(( x − 2 ), t: 8)$] #math.equation(block: false, alt: "a cubed b to the power 4")[$a^(3) b^(4)$] term of #math.equation(block: false, alt: "open parenthesis 2 a plus b close parenthesis to the power 7")[$attach(( 2 a + b ), t: 7)$] #solutionbox[ 280 ] === Practice Test In the following exercises, write the first five terms of the sequence whose general term is given. #math.equation(block: true, alt: "a sub n equals the fraction 5 n minus 3 over 3 to the power n")[$a_(n) = frac(5 n − 3, 3^(n))$] #math.equation(block: true, alt: "a sub n equals the fraction open parenthesis n plus 2 close parenthesis ! over open parenthesis n plus 3 close parenthesis !")[$a_(n) = frac(( n + 2 ) !, ( n + 3 ) "!")$] #solutionbox[ #math.equation(block: true, alt: "the fraction 1 over 4 , the fraction 1 over 5 , the fraction 1 over 6 , the fraction 1 over 7 , the fraction 1 over 8")[$frac(1, 4) , frac(1, 5) , frac(1, 6) , frac(1, 7) , frac(1, 8)$] ] Find a general term for the sequence, #math.equation(block: false, alt: "minus the fraction 2 over 3 , minus the fraction 4 over 5 , minus the fraction 6 over 7 , minus the fraction 8 over 9 , minus the fraction 10 over 11 , …")[$− frac(2, 3) , − frac(4, 5) , − frac(6, 7) , − frac(8, 9) , − frac(10, 11) , "…"$] Expand the partial sum and find its value. #math.equation(block: false, alt: "∑ i equals 14 open parenthesis −4 close parenthesis to the power i")[$display(limits(∑)^(4)_(i = 1) attach(( −4 ), t: i))$] #solutionbox[ #math.equation(block: true, alt: "−4 plus 16 minus 64 plus 256 equals 204")[$−4 + 16 − 64 + 256 = 204$] ] Write the following using summation notation. #math.equation(block: false, alt: "−1 plus the fraction 1 over 4 minus the fraction 1 over 9 plus the fraction 1 over 16 minus the fraction 1 over 25")[$−1 + frac(1, 4) − frac(1, 9) + frac(1, 16) − frac(1, 25)$] Write the first five terms of the arithmetic sequence with the given first term and common difference. #math.equation(block: false, alt: "a sub 1 equals −13")[$a_(1) = −13$] and #math.equation(block: false, alt: "d equals 3")[$d = 3$] #solutionbox[ #math.equation(block: true, alt: "−13 , −10 , −7 , −4 , −1")[$−13 , −10 , −7 , −4 , −1$] ] Find the twentieth term of an arithmetic sequence where the first term is two and the common difference is #math.equation(block: false, alt: "−7.")[$−7 .$] Find the twenty-third term of an arithmetic sequence whose seventh term is #math.equation(block: false, alt: "11")[$11$] and common difference is three. Then find a formula for the general term. #solutionbox[ #math.equation(block: false, alt: "a sub 23 equals 59.")[$a_(23) = 59 .$] The general term is #math.equation(block: false, alt: "a sub n equals 3 n minus 10.")[$a_(n) = 3 n − 10 .$] ] Find the first term and common difference of an arithmetic sequence whose ninth term is #math.equation(block: false, alt: "−1")[$−1$] and the sixteenth term is #math.equation(block: false, alt: "−15.")[$−15 .$] Then find a formula for the general term. Find the sum of the first 25 terms of the arithmetic sequence, #math.equation(block: false, alt: "5 , 9 , 13 , 17 , 21 , …")[$5 , 9 , 13 , 17 , 21 , "…"$] #solutionbox[ #math.equation(block: true, alt: "1,325")[$1,325$] ] Find the sum of the first 50 terms of the arithmetic sequence whose general term is #math.equation(block: false, alt: "a sub n equals −3 n plus 100.")[$a_(n) = −3 n + 100 .$] Find the sum. #math.equation(block: false, alt: "∑ i equals 140 open parenthesis 5 i minus 21 close parenthesis")[$display(limits(∑)^(40)_(i = 1) ( 5 i − 21) )$] #solutionbox[ #math.equation(block: true, alt: "3,260")[$3,260$] ] In the following exercises, determine if the sequence is arithmetic, geometric, or neither. If arithmetic, then find the common difference. If geometric, then find the common ratio. #math.equation(block: true, alt: "14 , 3 , −8 , −19 , −30 , −41 , …")[$14 , 3 , −8 , −19 , −30 , −41 , "…"$] #math.equation(block: true, alt: "324 , 108 , 36 , 12 , 4 , the fraction 4 over 3 , …")[$324 , 108 , 36 , 12 , 4 , frac(4, 3) , "…"$] #solutionbox[ The sequence is geometric with common ratio #math.equation(block: false, alt: "r equals the fraction 1 over 3.")[$r = frac(1, 3) .$] ] Write the first five terms of the geometric sequence with the given first term and common ratio. #math.equation(block: false, alt: "a sub 1 equals 6")[$a_(1) = 6$] and #math.equation(block: false, alt: "r equals −2")[$r = −2$] In the geometric sequence whose first term and common ratio are #math.equation(block: false, alt: "a sub 1 equals 5")[$a_(1) = 5$] and #math.equation(block: false, alt: "r equals 4 ,")[$r = 4 ,$] find #math.equation(block: false, alt: "a sub 11.")[$a_(11) .$] #solutionbox[ #math.equation(block: true, alt: "5,242,880")[$5,242,880$] ] Find #math.equation(block: false, alt: "a sub 10")[$a_(10)$] of the geometric sequence, #math.equation(block: false, alt: "1250 , 250 , 50 , 10 , 2 , the fraction 2 over 5 , … .")[$1250 , 250 , 50 , 10 , 2 , frac(2, 5) , "…" "."$] Then find a formula for the general term. Find the sum of the first thirteen terms of the geometric sequence, #math.equation(block: false, alt: "2 , −6 , 18 , −54 , 162 , −486 …")[$2 , −6 , 18 , −54 , 162 , −486 "…"$] #solutionbox[ #math.equation(block: true, alt: "797,162")[$797,162$] ] In the following exercises, find the sum. #math.equation(block: true, alt: "∑ i equals 195 open parenthesis 2 close parenthesis to the power i")[$display(limits(∑)^(9)_(i = 1) 5 attach(( 2 ), t: i))$] #math.equation(block: true, alt: "1 minus the fraction 1 over 5 plus the fraction 1 over 25 minus the fraction 1 over 125 plus the fraction 1 over 625 minus the fraction 1 over 3125 plus …")[$1 − frac(1, 5) + frac(1, 25) − frac(1, 125) + frac(1, 625) − frac(1, 3125) + "…"$] #solutionbox[ #math.equation(block: true, alt: "the fraction 5 over 6")[$frac(5, 6)$] ] Write the repeating decimal as a fraction. #math.equation(block: false, alt: "0.81 bar")[$0. accent(81, —)$] Dave just got his first full-time job after graduating from high school at age 18. He decided to invest \$450 per month in an IRA (an annuity). The interest on the annuity is 6% which is compounded monthly. How much will be in Adam’s account when he retires at his sixty-fifth birthday? #solutionbox[ #math.equation(block: true, alt: "$ 1,409,344.19")[$"$" 1,409,344.19$] ] Expand the binomial using Pascal’s Triangle. #math.equation(block: false, alt: "open parenthesis m minus 2 n close parenthesis to the power 5")[$attach(( m − 2 n ), t: 5)$] Evaluate each binomial coefficient. ⓐ #math.equation(block: false, alt: "open parenthesis 8; 1 close parenthesis")[$( 8 \ 1 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis 16; 16 close parenthesis")[$( 16 \ 16 )$] ⓒ #math.equation(block: false, alt: "open parenthesis 12; 0 close parenthesis")[$( 12 \ 0 )$] ⓓ #math.equation(block: false, alt: "open parenthesis 10; 6 close parenthesis")[$( 10 \ 6 )$] #solutionbox[ ⓐ 8 ⓑ 1 ⓒ 1 ⓓ 210 ] Expand the binomial using the Binomial Theorem. #math.equation(block: false, alt: "open parenthesis 4 x plus 5 y close parenthesis cubed")[$attach(( 4 x + 5 y ), t: 3)$]