#set document(title: "9.5 The Binomial Expansion", 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")) == 9.5#h(0.6em)The Binomial Expansion Earlier we studied products of polynomials, and in particular we found expanded forms for powers of binomials such as #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared")[$( a + b )^(2)$] and #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis cubed")[$( a + b )^(3)$]. The amount of work involved in expanding such powers increases as the exponent gets larger. In this section we will learn how to raise a binomial to any positive integer power, without having to perform the polynomial products. #notebox("Investigation", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Powers of Binomials] In this investigation we will look for patterns in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$]. We begin by computing several such powers. Expand each power and fill in the blanks. Arrange the terms in each expansion in descending powers of #math.equation(block: false, alt: "a")[$a$]. + #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 0 equals")[$( a + b )^(0) =$]\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ + #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 1 equals")[$( a + b )^(1) =$]\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ + #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared equals")[$( a + b )^(2) =$]\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ + #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis cubed equals")[$( a + b )^(3) =$]\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ #linebreak() Hint: Start by writing #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis cubed equals open parenthesis a plus b close parenthesis open parenthesis a plus b close parenthesis squared")[$( a + b )^(3) = ( a + b ) ( a + b )^(2)$] and use your answer to \#3. + #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 4 equals")[$( a + b )^(4) =$]\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ #linebreak() Hint: Start by writing #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 4 equals open parenthesis a plus b close parenthesis open parenthesis a plus b close parenthesis to the power 4")[$( a + b )^(4) = ( a + b ) ( a + b )^(4)$] and use your answer to \#4. + #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 5 equals")[$( a + b )^(5) =$]\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ + Do you see a relationship between the exponent #math.equation(block: false, alt: "n")[$n$] and the number of terms in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$]? (Notice that for #math.equation(block: false, alt: "n equals 0")[$n = 0$] we have #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 0 equals 1")[$( a + b )^(0) = 1$], which has one term.) Fill in the table below.#figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$" " n " "$]], [Number of terms #linebreak() in #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$] #linebreak()]), [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [#math.equation(block: false, alt: "1")[$1$]], [$#hide($0000$)$], [#math.equation(block: false, alt: "2")[$2$]], [$#hide($0000$)$], [#math.equation(block: false, alt: "3")[$3$]], [$#hide($0000$)$], [#math.equation(block: false, alt: "4")[$4$]], [$#hide($0000$)$], [#math.equation(block: false, alt: "5")[$5$]], [$#hide($0000$)$], )) + First observation: In general, the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$] has \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ terms. + Next we'll consider the exponents on #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] in each term of the expansions. Refer to your expanded powers in parts 1-5, and fill in the next table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$" " n " "$]], [First term of #linebreak() #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$] #linebreak()], [Last term of #linebreak() #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$] #linebreak()], [Sum of exponents #linebreak() on #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] #linebreak() in each term #linebreak()]), [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "1")[$1$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2")[$2$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "3")[$3$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "4")[$4$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "5")[$5$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Second observation: In any term of the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$], the sum of the exponents on #math.equation(block: false, alt: "a")[$a$] and on #math.equation(block: false, alt: "b")[$b$] is \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ + In fact, we can be more specific in describing the exponents in the expansions. We will use #math.equation(block: false, alt: "k")[$k$] to label the terms in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$], starting with #math.equation(block: false, alt: "k equals 0")[$k = 0$]. For example, for #math.equation(block: false, alt: "n equals 2")[$n = 2$] we label the terms as follows: . #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis squared equals, a squared plus 2 a b plus b squared; k equals 0 k equals 1 k equals 2")[$( a + b )^(2) = & " " " " a^(2) " " " " " " " " + " " " " " " " " 2 a b " " " " " " " " + " " " " " " " " b^(2) \ & k = 0 " " " " " " " " " " " " " " " " k = 1 " " " " " " " " " " " " " " " " k = 2$] We can make a table showing the exponents on #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] in each term of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared")[$( a + b )^(2)$]: #linebreak() Case #math.equation(block: false, alt: "n equals 2")[$n = 2$]:#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "k")[$k$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]]), [Exponent on #math.equation(block: false, alt: "a")[$a$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [Exponent on #math.equation(block: false, alt: "b")[$b$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], )) #linebreak() Complete the tables shown below for the cases #math.equation(block: false, alt: "n equals 3 , n equals 4")[$n = 3 , " " n = 4$] and #math.equation(block: false, alt: "n equals 5")[$n = 5$]. #linebreak() Case #math.equation(block: false, alt: "n equals 3")[$n = 3$]:#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "k")[$k$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]]), [Exponent on #math.equation(block: false, alt: "a")[$a$]], [#math.equation(block: false, alt: "3")[$3$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [Exponent on #math.equation(block: false, alt: "b")[$b$]], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #linebreak() Case #math.equation(block: false, alt: "n equals 4")[$n = 4$]:#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "k")[$k$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [Exponent on #math.equation(block: false, alt: "a")[$a$]], [#math.equation(block: false, alt: "4")[$4$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [Exponent on #math.equation(block: false, alt: "b")[$b$]], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #linebreak() Case #math.equation(block: false, alt: "n equals 5")[$n = 5$]:#figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "k")[$k$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]]), [Exponent on #math.equation(block: false, alt: "a")[$a$]], [#math.equation(block: false, alt: "4")[$4$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [Exponent on #math.equation(block: false, alt: "b")[$b$]], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Third observation: The variable factors of the #math.equation(block: false, alt: "k to the power th")[$k^("th")$] term in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$] may be expressed as \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. (Fill in the correct powers in terms of #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "n")[$n$] for #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$].) In the next investigation we will look for patterns in the coefficients of the terms of the expansions. ] === Powers of Other Binomials We can use what we learned from the Investigation to raise other binomials to powers. #examplebox("Example 1")[][ Expand each power of a binomial. + #math.equation(block: false, alt: "open parenthesis x plus 1 close parenthesis cubed")[$( x + 1 )^(3)$] + #math.equation(block: false, alt: "open parenthesis 2 m minus n close parenthesis to the power 4")[$( 2 m − n )^(4)$] #solutionbox[ + We know from the Investigation that #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis cubed equals a cubed plus 3 a squared b plus 3 a b squared plus b cubed")[$( a + b )^(3) = a^(3) + 3 a^(2) b + 3 a b^(2) + b^(3)$] We replace #math.equation(block: false, alt: "a")[$a$] with #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "b")[$b$] with 1 to get #math.equation(block: true, alt: "open parenthesis x plus 1 close parenthesis cubed, equals x cubed plus 3 x squared open parenthesis 1 close parenthesis plus 3 x open parenthesis 1 close parenthesis squared plus 1 cubed; equals x cubed plus 3 x squared plus 3 x plus 1")[$( x + 1 )^(3) & = x^(3) + 3 x^(2) ( 1 ) + 3 x ( 1 )^(2) + 1^(3) \ & = x^(3) + 3 x^(2) + 3 x + 1$] + This time we take the expansion #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 4 equals a to the power 4 plus 4 a cubed b plus 6 a squared b squared plus 4 a b cubed plus b to the power 4")[$( a + b )^(4) = a^(4) + 4 a^(3) b + 6 a^(2) b^(2) + 4 a b^(3) + b^(4)$] from the Investigation and let #math.equation(block: false, alt: "a equals 2 m")[$a = 2 m$] and #math.equation(block: false, alt: "b equals minus n")[$b = − n$]. #math.equation(block: true, alt: "open parenthesis 2 m minus n close parenthesis to the power 4, equals open parenthesis 2 m close parenthesis to the power 4 plus 4 open parenthesis 2 m close parenthesis cubed open parenthesis minus n close parenthesis plus 6 open parenthesis 2 m close parenthesis squared open parenthesis minus n close parenthesis squared plus; plus 4 open parenthesis 2 m close parenthesis open parenthesis minus n close parenthesis cubed plus open parenthesis minus n close parenthesis to the power 4; equals 16 m to the power 4 plus 4 open parenthesis 8 m cubed close parenthesis open parenthesis minus n close parenthesis plus 6 open parenthesis 4 m squared close parenthesis open parenthesis n squared close parenthesis plus; plus 4 open parenthesis 2 m close parenthesis open parenthesis minus n cubed close p")[$( 2 m − n )^(4) & = ( 2 m )^(4) + 4 ( 2 m )^(3) ( − n ) + 6 ( 2 m )^(2) ( − n )^(2) + \ & #h(2em) #h(2em) #h(2em) #h(2em) #h(2em) #h(1em) + 4 ( 2 m ) ( − n )^(3) + ( − n )^(4) \ & = 16 m^(4) + 4 ( 8 m^(3) ) ( − n ) + 6 ( 4 m^(2) ) ( n^(2) ) + \ & #h(2em) #h(2em) #h(2em) #h(2em) #h(2em) #h(1em) + 4 ( 2 m ) ( − n^(3) ) + n^(4) \ & = 16 m^(4) − 32 m^(3) n + 24 m^(2) n^(2) − 8 m n^(3) + n^(4)$] ] ] Expand #math.equation(block: false, alt: "open parenthesis r plus 2 s close parenthesis cubed equals")[$( r + 2 s )^(3) =$]\_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "r cubed plus 6 r squared s plus 12 r s squared plus 8 s cubed")[$r^(3) + 6 r^(2) s + 12 r s^(2) + 8 s^(3)$] ] Expand #math.equation(block: false, alt: "open parenthesis r plus 2 s close parenthesis cubed")[$( r + 2 s )^(3)$] #solutionbox[ #math.equation(block: true, alt: "r cubed plus 6 r squared s plus 12 r s squared plus 8 s cubed")[$r^(3) + 6 r^(2) s + 12 r s^(2) + 8 s^(3)$] ] === The Binomial Coefficient Now we continue our study of powers of binomials. What should the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 6")[$( a + b )^(6)$] look like? From the Iinvestigation we know that it should have 7 terms, with the exponents on #math.equation(block: false, alt: "a")[$a$] decreasing from 6 down to 0, and the exponents on #math.equation(block: false, alt: "b")[$b$] increasing from 0 up to 6. Thus, the expansion should have the form #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 6, equals bar a to the power 6 plus bar a to the power 5 b plus bar a to the power 4 b squared plus bar a cubed b cubed plus; plus bar a squared b to the power 4 plus bar a b to the power 5 plus bar b to the power 6")[$( a + b )^(6) & = underline(#h(1em)) a^(6) + underline(#h(1em)) a^(5) b + underline(#h(1em)) a^(4) b^(2) + underline(#h(1em)) a^(3) b^(3) + \ & #h(2em) #h(2em) #h(2em) #h(2em) #h(2em) #h(2em) + underline(#h(1em)) a^(2) b^(4) + underline(#h(1em)) a b^(5) + underline(#h(1em)) b^(6)$] But what about the numerical coefficients of the terms? We will use the notation #math.equation(block: false, alt: "n C sub k")[$n C_(k)$] for the coefficient of the #math.equation(block: false, alt: "k to the power th")[$k^("th")$] term in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$]. For example, in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 6")[$( a + b )^(6)$] above, - the coefficient of #math.equation(block: false, alt: "a to the power 6")[$a^(6)$] is denoted by #math.equation(block: false, alt: "6 C sub 0")[$6 C_(0)$], - the coefficient of #math.equation(block: false, alt: "a to the power 5 b to the power 1")[$a^(5) b^(1)$] is #math.equation(block: false, alt: "6 C sub 1")[$6 C_(1)$], - the coefficient of #math.equation(block: false, alt: "a to the power 4 b squared")[$a^(4) b^(2)$] is #math.equation(block: false, alt: "6 C sub 2")[$6 C_(2)$], and so on. Note that the 6 in front of the #math.equation(block: false, alt: "C")[$C$] indicates that #math.equation(block: false, alt: "n equals 6")[$n = 6$], and the number following #math.equation(block: false, alt: "C")[$C$] corresponds to the exponent on #math.equation(block: false, alt: "b")[$b$]. The symbol #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$] is called the #strong[binomial coefficient]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Binomial Coefficient] The #strong[binomial coefficient] #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$] is the coefficient of the term containing #math.equation(block: false, alt: "b to the power k")[$" " b^(k) " "$] in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$]. ] #examplebox("Example 2")[][ Use your expansions from the Investigation to evaluate the following binomial coefficients. + #math.equation(block: false, alt: "4 C sub 3")[$4 C_(3)$] + #math.equation(block: false, alt: "5 C sub 2")[$5 C_(2)$] #solutionbox[ + #math.equation(block: false, alt: "4 C sub 3")[$4 C_(3)$] is the coefficient of the term containing #math.equation(block: false, alt: "b cubed")[$b^(3)$] in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 4")[$( a + b )^(4)$]. Referring to Step 5 in the Investigation, we see that the coefficient of the term #math.equation(block: false, alt: "4 a b cubed")[$4 a b^(3)$] is 4. Thus, #math.equation(block: false, alt: "4 C sub 3 equals 4")[$4 C_(3) = 4$]. + #math.equation(block: false, alt: "5 C sub 2")[$5 C_(2)$] is the coefficient of the term containing #math.equation(block: false, alt: "b squared")[$b^(2)$] in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 5")[$( a + b )^(5)$]. Referring to Step 6 in the Investigation, we see that the coefficient of the term #math.equation(block: false, alt: "10 a cubed b squared")[$10 a^(3) b^(2)$] is 10. Thus, #math.equation(block: false, alt: "5 C sub 2 equals 10")[$5 C_(2) = 10$]. ] ] Evaluate the binomial coefficients. + #math.equation(block: false, alt: "5 C sub 4 equals")[$5 C_(4) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "5 C sub 0 equals")[$5 C_(0) =$]\_\_\_\_\_ #solutionbox[ + 5 + 1 ] Evaluate the binomial coefficients. + #math.equation(block: false, alt: "5 C sub 4")[$5 C_(4)$] + #math.equation(block: false, alt: "5 C sub 0")[$5 C_(0)$] #solutionbox[ + 5 + 1 ] #notebox("Investigation", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Pascal's Triangle] To get a clearer picture of the binomial coefficients, consider again the expansions of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$] you calculated in the last Investigation, but this time look only at the numerical coefficients of each term: #figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n equals 0")[$n = 0$]], [$#hide($0000$)$], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "n equals 1")[$n = 1$]], [], [#math.equation(block: false, alt: "11")[$1 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 2")[$n = 2$]], [], [#math.equation(block: false, alt: "121")[$1 #hide($000$) 2 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 3")[$n = 3$]], [], [#math.equation(block: false, alt: "1331")[$1 #hide($000$) 3 #hide($000$) 3 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 4")[$n = 4$]], [], [#math.equation(block: false, alt: "14641")[$1 #hide($000$) 4 #hide($000$) 6 #hide($000$) 4 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 5")[$n = 5$]], [], [#math.equation(block: false, alt: "15101051")[$1 #hide($000$) 5 #hide($i i 0$) 10 #hide($00$) 10 #hide($0 i i$) 5 #hide($000$) 1$]], )) This triangular array of numbers is known as #strong[Pascal's triangle]. It has many interesting and surprising properties that have been extensively studied. We might first make the following observations. + Each row of Pascal's triangle begins with the number \_\_\_\_\_\_ and ends with the number \_\_\_\_\_\_. + The second number and the next-to-last number in the #math.equation(block: false, alt: "n to the power th")[$n^("th")$] row are \_\_\_\_\_. #linebreak() There is an interesting pattern for the rest of the numbers in each row. Pick any number in the row #math.equation(block: false, alt: "n equals 4")[$n = 4$], and look at the two closest numbers in the previous row. (For example, if you picked 6, the two closest numbers in the previous row are 3 and 3.) Do you see a relationship between the numbers? Try the same thing with several numbers in row #math.equation(block: false, alt: "n equals 5")[$n = 5$] to test your theory. + Starting with the row #math.equation(block: false, alt: "n equals 2")[$n = 2$], any number in the triangle (except the first and last 1's in each row) can be found by \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. + Using your answer to question \#3, continue Pascal's triangle to include the row for #math.equation(block: false, alt: "n equals 6")[$n = 6$]. #linebreak() The numbers in Pascal's triangle are the binomial coefficients we have been looking for. Specifically, the number in the #math.equation(block: false, alt: "k to the power th")[$k^("th")$] position (starting with #math.equation(block: false, alt: "k equals 0")[$k = 0$]) of the #math.equation(block: false, alt: "n to the power th")[$n^("th")$] row of the triangle is #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$]. Thus, we can use the numbers in Pascal's triangle to expand #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$]. + Use Pascal's triangle to find the binomial coefficient #math.equation(block: false, alt: "6 C sub 4")[$attach(" ", b: 6) C_(4) " "$]. + Expand: #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 6")[$( a + b )^(6)$]: \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ + Expand: #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis to the power 6")[$( x − 2 )^(6)$]: \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. (Hint: Use your answer to \#6, replacing #math.equation(block: false, alt: "a")[$a$] by #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "b")[$b$] by #math.equation(block: false, alt: "minus 2")[$− 2$].) + Continue Pascal's triangle to include the row for #math.equation(block: false, alt: "n equals 7")[$n = 7$]. ] === Using Pascal's Triangle We can now write the expanded form of any binomial power without having to perform the multiplication. #examplebox("Example 3")[][ Find the expanded form of #math.equation(block: false, alt: "open parenthesis 3 r minus q close parenthesis to the power 6")[$( 3 r − q )^(6)$] without performing the multiplication. #solutionbox[ #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 6, equals a to the power 6 plus bar a to the power 5 b plus bar a to the power 4 b squared plus bar a cubed b cubed plus; plus bar a squared b to the power 4 plus bar a b to the power 5 plus b to the power 6")[$( a + b )^(6) & = a^(6) + underline(#h(1em)) a^(5) b + underline(#h(1em)) a^(4) b^(2) + underline(#h(1em)) a^(3) b^(3) + \ & #h(2em) #h(2em) #h(2em) #h(2em) #h(2em) #h(2em) #hide($+$) + underline(#h(1em)) a^(2) b^(4) + underline(#h(1em)) a b^(5) + b^(6)$] Next, fill in the blanks with the binomial coefficients from Pascal's triangle. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 6 equals a to the power 6 plus 6 a to the power 5 b plus 15 a to the power 4 b squared plus 20 a cubed b cubed plus 15 a squared b to the power 4 plus 6 a b to the power 5 plus b to the power 6")[$( a + b )^(6) = a^(6) + 6 a^(5) b + 15 a^(4) b^(2) + 20 a^(3) b^(3) + 15 a^(2) b^(4) + 6 a b^(5) + b^(6)$] Finally, replace #math.equation(block: false, alt: "a")[$a$] by #math.equation(block: false, alt: "3 r")[$3 r$] and replace #math.equation(block: false, alt: "b")[$b$] by #math.equation(block: false, alt: "minus q")[$− q$], then simplify to find #math.equation(block: true, alt: "open parenthesis 3 r minus q close parenthesis to the power 6, equals open parenthesis 3 r close parenthesis to the power 6 plus 6 open parenthesis 3 r close parenthesis to the power 5 open parenthesis minus q close parenthesis plus 15 open parenthesis 3 r close parenthesis to the power 4 open parenthesis minus q close parenthesis squared plus 20 open parenthesis 3 r close parenthesis cubed open parenthesis minus q close parenthesis cubed plus; plus 15 open parenthesis 3 r close parenthesis squared open parenthesis minus 6 close parenthesis to the power 4 plus 6 open parenthesis 3 r close parenthesis open parenthesis minus q close parenthesis to the power 5 plus open parenthesis minus q close parenthesis to the power 6; equals 729 r to the power 6 minus 1458 r to the power 5 q plus 1215 r")[$( 3 r − q )^(6) & = ( 3 r )^(6) + 6 ( 3 r )^(5) ( − q ) + 15 ( 3 r )^(4) ( − q )^(2) + 20 ( 3 r )^(3) ( − q )^(3) + \ & #h(2em) #h(2em) #h(2em) #h(2em) #h(1em) + 15 ( 3 r )^(2) ( − 6 )^(4) + 6 ( 3 r ) ( − q )^(5) + ( − q )^(6) \ & = 729 r^(6) − 1458 r^(5) q + 1215 r^(4) q^(2) − 540 r^(3) q^(3) + \ & #h(2em) #h(2em) #h(2em) #h(2em) #h(1em) + 135 r^(2) q^(2) − 18 r q^(5) + q^(6)$] ] ] Expand #math.equation(block: false, alt: "open parenthesis w plus 10 z close parenthesis to the power 6")[$( w + 10 z )^(6)$]. \_\_\_\_\_ Do not enter any commas. For example, use "10000" rather than "10,000". #solutionbox[ #math.equation(block: true, alt: "w to the power 6 plus 60 w to the power 5 z plus 1500 w to the power 4 z squared plus 20 , 000 w cubed z cubed, plus 150 , 000 w squared z to the power 4 plus; plus 600 , 000 w z to the power 5 plus 1 , 000 , 000 z to the power 6")[$w^(6) + 60 w^(5) z + 1500 w^(4) z^(2) + 20 , 000 w^(3) z^(3) & + 150 , 000 w^(2) z^(4) + \ & + 600 , 000 w z^(5) + 1 , 000 , 000 z^(6)$] ] Expand #math.equation(block: false, alt: "open parenthesis w plus 10 z close parenthesis to the power 6")[$( w + 10 z )^(6)$]. #solutionbox[ #math.equation(block: true, alt: "w to the power 6 plus 60 w to the power 5 z plus 1500 w to the power 4 z squared plus 20 , 000 w cubed z cubed, plus 150 , 000 w squared z to the power 4 plus; plus 600 , 000 w z to the power 5 plus 1 , 000 , 000 z to the power 6")[$w^(6) + 60 w^(5) z + 1500 w^(4) z^(2) + 20 , 000 w^(3) z^(3) & + 150 , 000 w^(2) z^(4) + \ & + 600 , 000 w z^(5) + 1 , 000 , 000 z^(6)$] ] === Factorial Notation Finding binomial coefficients from Pascal's triangle is tedious for large values of #math.equation(block: false, alt: "n")[$n$]. It would be convenient to have a formula for calculating the coefficients directly. For this formula we need a new symbol, #strong[#math.equation(block: false, alt: "n !")[$n !$]] (read #strong["#math.equation(block: false, alt: "n")[$n$] factorial"]), which is defined as follows: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Factorial] If #math.equation(block: false, alt: "n")[$n$] is a positive integer, #math.equation(block: true, alt: "n ! equals n times open parenthesis n minus 1 close parenthesis times open parenthesis n minus 2 close parenthesis times ⋯ times 3 times 2 times 1")[$n ! = n ⋅ ( n − 1 ) ⋅ ( n − 2 ) ⋅ ⋯ ⋅ 3 ⋅ 2 ⋅ 1$] ] For example, #math.equation(block: true, alt: "6 ! equals 5 times 4 times 3 times 2 times 1 equals 720 and 4 ! equals 4 times 3 times 2 times 1 equals 24")[$6 ! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720 " " " " " " " " " " "and" " " " " " " " " " " 4 ! = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The factorial symbol applies only to the variable or number it follows; for example, #math.equation(block: false, alt: "3 times 4 !")[$3 ⋅ 4 !$] is not equal to #math.equation(block: false, alt: "open parenthesis 3 times 4 close parenthesis !")[$( 3 ⋅ 4 ) !$]. ] #examplebox("Example 4")[][ Write each expression in expanded form and simplify. + #math.equation(block: false, alt: "2 n !")[$2 n !$] for #math.equation(block: false, alt: "n equals 4")[$n = 4$] + #math.equation(block: false, alt: "open parenthesis 2 n minus 1 close parenthesis !")[$( 2 n − 1 ) !$] for #math.equation(block: false, alt: "n equals 4")[$n = 4$] #solutionbox[ + #math.equation(block: false, alt: "2 n ! equals 2 times 4 ! equals 2 open parenthesis 4 times 3 times 2 times 1 close parenthesis equals 48")[$2 n ! = 2 ⋅ 4 ! = 2 ( 4 ⋅ 3 ⋅ 2 ⋅ 1 ) = 48$] + #math.equation(block: false, alt: "open parenthesis 2 n minus 1 close parenthesis ! equals 7 ! equals 7 times 6 times 5 times 4 times 3 times 2 times 1 equals 5040")[$( 2 n − 1 ) ! = 7 ! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5040$] ] ] Evalute #math.equation(block: false, alt: "4 times 3 ! equals")[$4 ⋅ 3 ! =$]\_\_\_\_\_ #solutionbox[ 24 ] Evalute #math.equation(block: false, alt: "4 times 3 !")[$4 ⋅ 3 !$] #solutionbox[ 24 ] #examplebox("Example 5")[][ Write #math.equation(block: false, alt: "open parenthesis 3 n plus 1 close parenthesis !")[$( 3 n + 1 ) !$] in factored form, showing the first three factors and the last three factors. #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 n plus 1 close parenthesis ! equals open parenthesis 3 n plus 1 close parenthesis times open parenthesis 3 n close parenthesis times open parenthesis 3 n minus 1 close parenthesis times ⋯ times 3 times 2 times 1")[$( 3 n + 1 ) ! = ( 3 n + 1 ) ⋅ ( 3 n ) ⋅ ( 3 n − 1 ) ⋅ ⋯ ⋅ 3 ⋅ 2 ⋅ 1$] ] ] Write the first three factors of + #math.equation(block: false, alt: "2 n !")[$2 n !$] #linebreak() \_\_\_\_\_ + #math.equation(block: false, alt: "open parenthesis 2 n close parenthesis !")[$( 2 n ) !$] #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "2 times n times open parenthesis n minus 1 close parenthesis")[$2 ⋅ n ⋅ ( n − 1 )$] + #math.equation(block: false, alt: "open parenthesis 2 n close parenthesis times open parenthesis 2 n minus 1 close parenthesis times open parenthesis 2 n minus 2 close parenthesis")[$( 2 n ) ⋅ ( 2 n − 1 ) ⋅ ( 2 n − 2 )$] ] Write the first three factors of + #math.equation(block: false, alt: "2 n !")[$2 n !$] + #math.equation(block: false, alt: "open parenthesis 2 n close parenthesis !")[$( 2 n ) !$] #solutionbox[ + #math.equation(block: false, alt: "2 times n times open parenthesis n minus 1 close parenthesis")[$2 ⋅ n ⋅ ( n − 1 )$] + #math.equation(block: false, alt: "open parenthesis 2 n close parenthesis times open parenthesis 2 n minus 1 close parenthesis times open parenthesis 2 n minus 2 close parenthesis")[$( 2 n ) ⋅ ( 2 n − 1 ) ⋅ ( 2 n − 2 )$] ] Note that #math.equation(block: false, alt: "7 ! equals 7 times 6 times 5 times 4 times 3 times 2 times 1")[$7 ! = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1$] can be written as #math.equation(block: true, alt: "7 times 6 ! equals 7 times open parenthesis 6 times 5 times 4 times 3 times 2 times 1 close parenthesis")[$7 ⋅ 6 ! = 7 ⋅ ( 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 )$] and #math.equation(block: false, alt: "5 ! equals 5 times 4 times 3 times 2 times 1")[$5 ! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1$] can be written as #math.equation(block: true, alt: "5 times 4 ! equals 5 times open parenthesis 4 times 3 times 2 times 1 close parenthesis")[$5 ⋅ 4 ! = 5 ⋅ ( 4 ⋅ 3 ⋅ 2 ⋅ 1 )$] In general, we have the following relationship: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "n ! equals n times open parenthesis n minus 1 close parenthesis !")[$n ! = n ⋅ ( n − 1 ) !$] ] This relationship can be helpful in simplifying expressions involving factorials. #examplebox("Example 6")[][ Write each expression in expanded form and simplify. + #math.equation(block: false, alt: "the fraction 6 ! over 3 !")[$display(frac(6 !, 3 !))$] + #math.equation(block: false, alt: "the fraction 4 ! 6 ! over 8 !")[$display(frac(4 ! " " 6 !, 8 !))$] #solutionbox[ + #math.equation(block: false, alt: "the fraction 6 ! over 3 ! equals the fraction 6 times 5 times 4 times 3 ! over 3 ! equals 6 times 5 times 4 equals 120")[$display(frac(6 !, 3 !)) = display(frac(6 ⋅ 5 ⋅ 4 ⋅ 3 !, 3 !)) = 6 ⋅ 5 ⋅ 4 = 120$] + #math.equation(block: false, alt: "the fraction 4 ! 6 ! over 8 ! equals the fraction 4 times 3 times 2 times 1 times 6 ! over 8 times 7 times 6 ! equals the fraction 3 over 7")[$display(frac(4 ! " " 6 !, 8 !)) = display(frac(4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 6 !, 8 ⋅ 7 ⋅ 6 !)) = display(frac(3, 7))$] ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the previous Example, note that #math.equation(block: false, alt: "the fraction 6 ! over 3 !")[$display(frac(6 !, 3 !))$] is not equal to #math.equation(block: false, alt: "2 !")[$2 !$]. ] Write #math.equation(block: false, alt: "the fraction 5 ! over 3 ! 2 !")[$display(frac(5 !, 3 ! " " 2 !))$] in expanded form and simplify. Simplified form: \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "10")[$10$] ] Write #math.equation(block: false, alt: "the fraction 5 ! over 3 ! 2 !")[$display(frac(5 !, 3 ! " " 2 !))$] in expanded form and simplify. #solutionbox[ #math.equation(block: true, alt: "10")[$10$] ] We have defined #math.equation(block: false, alt: "n !")[$n !$] when #math.equation(block: false, alt: "n")[$n$] is a positive integer. What if #math.equation(block: false, alt: "n equals 0")[$n = 0$]? To be consistent with the equation #math.equation(block: false, alt: "n ! equals n times open parenthesis n minus 1 close parenthesis !")[$n ! = n ⋅ ( n − 1 ) !$] we must have #math.equation(block: true, alt: "1 ! equals 1 times open parenthesis 1 minus 1 close parenthesis ! or 1 ! equals 1 times open parenthesis 0 close parenthesis !")[$1 ! = 1 ⋅ ( 1 − 1 ) ! " " " " " " " " " " " " " " " " "or" " " " " " " " " " " " " " " " " 1 ! = 1 ⋅ ( 0 ) !$] This leads us to the following definition. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Zero Factorial] #math.equation(block: true, alt: "0 ! equals 1")[$0 ! = 1$] ] Note that both #math.equation(block: false, alt: "1 !")[$1 !$] and #math.equation(block: false, alt: "0 !")[$0 !$] are equal to #math.equation(block: false, alt: "1")[$1$]. === The Binomial Coefficient in Factorial Notation Earlier we introduced the notation #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$] for the numerical coefficients in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$]. These binomial coefficients, which are given by Pascal's triangle, can also be expressed using factorial notation as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Binomial Coefficients] For #math.equation(block: false, alt: "n greater than or equal to 0")[$n ≥ 0$] and #math.equation(block: false, alt: "0 less than or equal to k less than or equal to n")[$0 ≤ k ≤ n$], #math.equation(block: true, alt: "n C sub k equals the fraction n ! over open parenthesis n minus k close parenthesis ! k !")[$attach(" ", b: n) C_(k) " " = display(frac(n !, ( n − k ) ! " " k !))$] ] #examplebox("Example 7")[][ Evaluate each binomial coefficient. + #math.equation(block: false, alt: "6 C sub 2")[$attach(" ", b: 6) C_(2) " "$] + #math.equation(block: false, alt: "9 C sub 8")[$attach(" ", b: 9) C_(8) " "$] #solutionbox[ + #math.equation(block: false, alt: "6 C sub 2 equals the fraction 6 ! over open parenthesis 6 minus 2 close parenthesis ! 2 ! equals the fraction 6 ! over 4 ! 2 ! equals the fraction 6 times 5 times 4 ! over 4 ! times 2 times 1 equals 15")[$attach(" ", b: 6) C_(2) " " = display(frac(6 !, ( 6 − 2 ) ! " " 2 !)) = display(frac(6 !, 4 ! " " 2 !)) = display(frac(6 ⋅ 5 ⋅ 4 !, 4 ! " " ⋅ 2 ⋅ 1)) = 15$] + #math.equation(block: false, alt: "9 C sub 8 equals the fraction 9 ! over open parenthesis 9 minus 8 close parenthesis ! 8 ! equals the fraction 9 ! over 1 ! 8 ! equals the fraction 9 times 8 ! over 1 times 8 ! equals 9")[$attach(" ", b: 9) C_(8) " " = display(frac(9 !, ( 9 − 8 ) ! " " 8 !)) = display(frac(9 !, 1 ! " " 8 !)) = display(frac(9 ⋅ 8 !, 1 ⋅ 8 !)) = 9$] ] ] Evaluate the binomial coefficient #math.equation(block: false, alt: "8 C sub 6 equals")[$#h(1em)_(8) C_(6) =$] \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "10")[$10$] ] Evaluate the binomial coefficient #math.equation(block: false, alt: "8 C sub 6")[$attach(" ", b: 8) C_(6)$]. #solutionbox[ #math.equation(block: true, alt: "10")[$10$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ You can check that the formula works for #math.equation(block: false, alt: "k equals 0")[$k = 0$] when we use #math.equation(block: false, alt: "0 ! equals 1")[$0 ! = 1$]. You should find that #math.equation(block: true, alt: "n C sub 0 equals the fraction n ! over open parenthesis n minus 0 close parenthesis ! 0 ! equals the fraction n ! over n ! open parenthesis 1 close parenthesis equals 1")[$attach(" ", b: n) C_(0) " " = display(frac(n !, ( n − 0 ) ! " " 0 !)) = display(frac(n !, n ! " " ( 1 ))) = 1$] so the formula correctly gives us the coefficient of #math.equation(block: false, alt: "a to the power n")[$a^(n)$] in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$]. ] We now have two methods for computing the binomial coefficient #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$] : Pascal's triangle and our formula using factorials. When #math.equation(block: false, alt: "n")[$n$] is small, especially if all the coefficients of the binomial expansion are needed, Pascal's triangle is often the easier choice. But when #math.equation(block: false, alt: "n")[$n$] is large or if only one coefficient is required, the factorial formula for #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$] is probably quicker than Pascal's triangle. #examplebox("Example 8")[][ Find the coefficient of #math.equation(block: false, alt: "m to the power 11 n cubed")[$m^(11) n^(3)$] in the expansion of #math.equation(block: false, alt: "open parenthesis m plus n close parenthesis to the power 14")[$( m + n )^(14)$]. #solutionbox[ We are simply replacing #math.equation(block: false, alt: "a")[$a$] with #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "b")[$b$] with #math.equation(block: false, alt: "n")[$n$] in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 14")[$( a + b )^(14)$], so the term containing #math.equation(block: false, alt: "m to the power 11 n cubed")[$m^(11) n^(3)$] has #math.equation(block: false, alt: "n equals 14")[$n = 14$] and #math.equation(block: false, alt: "k equals 3")[$k = 3$], and its coefficient is #math.equation(block: false, alt: "14 C sub 3")[$attach(" ", b: 14) C_(3) " "$]. #math.equation(block: true, alt: "14 C sub 3 equals the fraction 14 ! over open parenthesis 14 minus 3 close parenthesis ! 3 ! equals the fraction 14 times 13 times 12 times 11 ! over 11 ! 3 ! equals the fraction 14 times 13 times 12 over 3 times 2 times 1 equals 364")[$attach(" ", b: 14) C_(3) " " = display(frac(14 !, ( 14 − 3 ) ! " " 3 !)) = display(frac(14 ⋅ 13 ⋅ 12 ⋅ 11 !, 11 ! " " 3 !)) = display(frac(14 ⋅ 13 ⋅ 12, 3 ⋅ 2 ⋅ 1)) = 364$] The coefficient of #math.equation(block: false, alt: "m to the power 11 n cubed")[$m^(11) n^(3)$] is #math.equation(block: false, alt: "364")[$364$]. ] ] Find the coefficient of #math.equation(block: false, alt: "s cubed t to the power 4")[$s^(3) t^(4)$] in the expansion of #math.equation(block: false, alt: "open parenthesis s plus t close parenthesis to the power 7")[$( s + t )^(7)$]. Answer: \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "35")[$35$] ] Find the coefficient of #math.equation(block: false, alt: "s cubed t to the power 4")[$s^(3) t^(4)$] in the expansion of #math.equation(block: false, alt: "open parenthesis s plus t close parenthesis to the power 7")[$( s + t )^(7)$]. #solutionbox[ #math.equation(block: true, alt: "35")[$35$] ] === The Binomial Theorem Consider the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 6")[$( a + b )^(6)$] written with the #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$] notation for the coefficients. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 6, equals a to the power 6 b to the power 0 plus 6 C sub 1 a to the power 5 b to the power 1 plus 6 C sub 2 a to the power 4 b squared plus 6 C sub 3 a cubed b cubed plus 6 C sub 4 a squared b to the power 4 plus 6 C sub 5 a to the power 1 b to the power 5 plus 6 C sub 6 a to the power 0 b to the power 6; k equals 0 k equals 1 k equals 2 k equals 3 k equals 4 k equals 5 k equals 6")[$( a + b )^(6) & = a^(6) b^(0) + 6 C_(1) " " a^(5) b^(1) + 6 C_(2) " " a^(4) b^(2) + 6 C_(3) " " a^(3) b^(3) + 6 C_(4) " " a^(2) b^(4) + 6 C_(5) " " a^(1) b^(5) + 6 C_(6) " " a^(0) b^(6) \ & #h(1em) k = 0 " " " " " " " " " " " " " " k = 1 " " " " " " " " " " " " " " " " " " " " " " " " k = 2 " " " " " " " " " " " " " " " " " " " " " " " " k = 3 " " " " " " " " " " " " " " " " " " " " " " " " k = 4 " " " " " " " " " " " " " " " " " " " " " " " " k = 5 " " " " " " " " " " " " " " " " " " " " " " k = 6$] Check that each term can be written in the form #math.equation(block: true, alt: "6 C sub k a to the power 6 minus k b to the power k")[$6 C_(k) " " a^(6 − k) b^(k)$] for #math.equation(block: false, alt: "k equals 0")[$k = 0$] to #math.equation(block: false, alt: "k equals 6")[$k = 6$]. This means that we can use sigma notation to write the sum of the terms in the expansion. #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power 6 equals ∑ k equals 066 C sub k a to the power 6 minus k b to the power k")[$( a + b )^(6) = limits(∑)^(6)_(k = 0) attach(" ", b: 6) C_(k) " " a^(6 − k) b^(k)$] Notice that the sigma notation indicates terms for #math.equation(block: false, alt: "k equals 0")[$k = 0$] to #math.equation(block: false, alt: "k equals 6")[$k = 6$], so there are 7 terms in the expansion, as there should be. Also note that the sum of the exponents on #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] is #math.equation(block: false, alt: "open parenthesis 6 minus k close parenthesis plus k equals 6")[$( 6 − k ) + k = 6$]. The equation above is a special case of the #strong[binomial theorem], which uses everything we have learned to write the expanded form of a power of a binomial in the most compact form possible. We can write the general form of the theorem, for positive integers #math.equation(block: false, alt: "n")[$n$], as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Binomial Theorem] #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals ∑ k equals 0 n n C sub k a to the power n minus k b to the power k")[$( a + b )^(n) = " " " " display(limits(∑)^(n)_(k = 0) attach(" ", b: n) C_(k) " " a^(n − k) b^(k))$] ] #examplebox("Example 9")[][ + Use sigma notation to write the expanded form for #math.equation(block: false, alt: "open parenthesis x minus 2 y close parenthesis to the power 10")[$( x − 2 y )^(10)$]. + Find the term containing #math.equation(block: false, alt: "y to the power 7")[$y^(7)$], and simplify. #solutionbox[ + We apply the binomial theorem with #math.equation(block: false, alt: "n equals 10")[$n = 10$], replacing #math.equation(block: false, alt: "a")[$a$] by #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "b")[$b$] by #math.equation(block: false, alt: "minus 2 y")[$− 2 y$], to get #math.equation(block: true, alt: "open parenthesis x minus 2 y close parenthesis to the power 10 equals ∑ k equals 01010 C sub k x to the power 10 minus k open parenthesis minus 2 y close parenthesis to the power k")[$( x − 2 y )^(10) = " " " " display(limits(∑)^(10)_(k = 0) attach(" ", b: 10) C_(k) " " x^(10 − k) ( − 2 y )^(k))$] + The term containing #math.equation(block: false, alt: "y to the power 7")[$y^(7)$] corresponds to #math.equation(block: false, alt: "k equals 7")[$k = 7$]. The #math.equation(block: false, alt: "k equals 7")[$k = 7$] term looks like #math.equation(block: true, alt: "10 C sub 7 x to the power 10 minus 7 open parenthesis minus 2 y close parenthesis to the power 7 or 10 C sub 7 x cubed open parenthesis minus 2 y close parenthesis to the power 7")[$attach(" ", b: 10) C_(7) " " x^(10 − 7) ( − 2 y )^(7) " " " " " " " " "or" " " " " " " " " attach(" ", b: 10) C_(7) " " x^(3) ( − 2 y )^(7)$] The value of #math.equation(block: false, alt: "10 C sub 7")[$attach(" ", b: 10) C_(7) " "$] is #math.equation(block: true, alt: "10 C sub 7 equals the fraction 10 ! over open parenthesis 10 minus 7 close parenthesis ! 7 ! equals the fraction 10 times 9 times 8 times 7 ! over 3 ! 7 ! equals the fraction 10 times 9 times 8 over 3 times 2 times 1 equals 120")[$attach(" ", b: 10) C_(7) " " = display(frac(10 !, ( 10 − 7 ) ! " " 7 !)) = display(frac(10 ⋅ 9 ⋅ 8 ⋅ 7 !, 3 ! " " 7 !)) = display(frac(10 ⋅ 9 ⋅ 8, 3 ⋅ 2 ⋅ 1)) = 120$] and #math.equation(block: false, alt: "open parenthesis minus 2 y close parenthesis to the power 7 equals minus 128 y to the power 7")[$( − 2 y )^(7) = − 128 y^(7)$]. Thus, the term we want is #math.equation(block: true, alt: "10 C sub 7 x cubed open parenthesis minus 2 y close parenthesis to the power 7 equals 120 x cubed open parenthesis minus 128 y to the power 7 close parenthesis equals minus 15 , 360 x cubed y to the power 7")[$attach(" ", b: 10) C_(7) " " x^(3) ( − 2 y )^(7) = 120 x^(3) ( − 128 y^(7) ) = − 15 , 360 x^(3) y^(7)$] ] ] + Use sigma notation to write the expanded form for #math.equation(block: false, alt: "open parenthesis 2 m minus 3 n close parenthesis to the power 7")[$( 2 m − 3 n )^(7)$]. #linebreak() \_\_\_\_\_ + Find the term containing #math.equation(block: false, alt: "m cubed")[$m^(3)$], and simplify. #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "open parenthesis 2 m minus 3 n close parenthesis to the power 7 equals ∑ k equals 077 C sub k open parenthesis 2 m close parenthesis to the power 7 minus k open parenthesis minus 3 n close parenthesis to the power k")[$( 2 m − 3 n )^(7) = display(limits(∑)^(7)_(k = 0) attach(" ", b: 7) C_(k) " " ( 2 m )^(7 − k) ( − 3 n )^(k))$] + #math.equation(block: false, alt: "7 C sub 4 open parenthesis 2 m close parenthesis cubed open parenthesis minus 3 n close parenthesis to the power 4 equals 22 , 680 m cubed n to the power 4")[$attach(" ", b: 7) C_(4) " " ( 2 m )^(3) ( − 3 n )^(4) = 22 , 680 m^(3) n^(4)$] ] + Use sigma notation to write the expanded form for #math.equation(block: false, alt: "open parenthesis 2 m minus 3 n close parenthesis to the power 7")[$( 2 m − 3 n )^(7)$]. + Find the term containing #math.equation(block: false, alt: "m cubed")[$m^(3)$], and simplify. #solutionbox[ + #math.equation(block: false, alt: "open parenthesis 2 m minus 3 n close parenthesis to the power 7 equals ∑ k equals 077 C sub k open parenthesis 2 m close parenthesis to the power 7 minus k open parenthesis minus 3 n close parenthesis to the power k")[$( 2 m − 3 n )^(7) = display(limits(∑)^(7)_(k = 0) attach(" ", b: 7) C_(k) " " ( 2 m )^(7 − k) ( − 3 n )^(k))$] + #math.equation(block: false, alt: "7 C sub 4 open parenthesis 2 m close parenthesis cubed open parenthesis minus 3 n close parenthesis to the power 4 equals 22 , 680 m cubed n to the power 4")[$attach(" ", b: 7) C_(4) " " ( 2 m )^(3) ( − 3 n )^(4) = 22 , 680 m^(3) n^(4)$] ] === Section Summary ==== Vocabulary - Binomial coefficient - Pascal's triangle - Factorial ==== CONCEPTS + The binomial coefficient #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$] is the coefficient of the term containing #math.equation(block: false, alt: "b to the power k")[$b^(k)$] in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power n")[$( a + b )^(n)$]. + The numbers in Pascal's triangle are the binomial coefficients. Specifically, the number in the #math.equation(block: false, alt: "k to the power th")[$k^("th")$] position (starting with #math.equation(block: false, alt: "k equals 0")[$k = 0$]) of the #math.equation(block: false, alt: "n to the power th")[$n^("th")$] row of the triangle is #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$]. + If #math.equation(block: false, alt: "n")[$n$] is a positive integer, #math.equation(block: true, alt: "n ! equals n times open parenthesis n minus 1 close parenthesis times open parenthesis n minus 2 close parenthesis times ⋯ times 3 times 2 times 1")[$n ! = n ⋅ ( n − 1 ) ⋅ ( n − 2 ) ⋅ ⋯ ⋅ 3 ⋅ 2 ⋅ 1$] + For #math.equation(block: false, alt: "n greater than or equal to 0")[$n ≥ 0$] and #math.equation(block: false, alt: "0 less than or equal to k less than or equal to n")[$0 ≤ k ≤ n$], #math.equation(block: true, alt: "n C sub k equals the fraction n ! over open parenthesis n minus k close parenthesis ! k !")[$attach(" ", b: n) C_(k) " " = display(frac(n !, ( n − k ) ! " " k !))$] + The Binomial Theorem #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis to the power n equals ∑ k equals 0 n n C sub k a to the power n minus k b to the power k")[$( a + b )^(n) = " " " " display(limits(∑)^(n)_(k = 0) attach(" ", b: n) C_(k) " " a^(n − k) b^(k))$] ==== STUDY QUESTIONS + What does the symbol #math.equation(block: false, alt: "n C sub k")[$attach(" ", b: n) C_(k) " "$] denote? + Explain how to use Pascal's triangle when expanding a power of a binomial. + What is factorial notation? + State a formula for the binomial coefficients in factorial notation. + Explain how to use the binomial theorem to expand #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 7")[$( a + b )^(7)$]. ==== SKILLS Practice each skill in the Homework Problems listed. + Expand powers of binomials: \#1–22, 33–34 + Simplify factorial notation: \#23–24 + Evaluate binomial coefficients: \#25–26 + Find specific terms of binomial expansions: \#28–32, \#35-42 + Use Pascal's triangle: \#43–44 === Homework 9.5 For Problems 1–4, answer the questions without expanding the power. How many terms are in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 50")[$( a + b )^(50)$]? In the expansion of #math.equation(block: false, alt: "open parenthesis 2 x plus 3 y close parenthesis to the power 100")[$( 2 x + 3 y )^(100)$]? #math.equation(block: true, alt: "51 ; 101")[$51 ; 101$] How many terms are in the expansion of #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis to the power 75")[$( a + b )^(75)$]? In the expansion of #math.equation(block: false, alt: "open parenthesis 5 x minus 7 y close parenthesis to the power 200")[$( 5 x − 7 y )^(200)$]? What is the sum of the exponents on #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in each term of the expansion of #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power 100")[$( x + y )^(100)$]? In the expansion of #math.equation(block: false, alt: "open parenthesis 8 x minus 7 y close parenthesis to the power 50")[$( 8 x − 7 y )^(50)$]? #math.equation(block: true, alt: "100 ; 50")[$100 ; 50$] What is the sum of the exponents on #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in each term of the expansion of #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis to the power 200")[$( x + y )^(200)$]? In the expansion of #math.equation(block: false, alt: "open parenthesis 9 x plus 4 y close parenthesis to the power 75")[$( 9 x + 4 y )^(75)$]? Write down the portion of Pascal's triangle corresponding to rows from #math.equation(block: false, alt: "n equals 0")[$n = 0$] to #math.equation(block: false, alt: "n equals 10")[$n = 10$]. How many rows are involved? 11 rows #figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n equals 0")[$n = 0$]], [$#hide($0000$)$], [#math.equation(block: false, alt: "1")[$1$]]), [#math.equation(block: false, alt: "n equals 1")[$n = 1$]], [], [#math.equation(block: false, alt: "11")[$1 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 2")[$n = 2$]], [], [#math.equation(block: false, alt: "121")[$1 #hide($000$) 2 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 3")[$n = 3$]], [], [#math.equation(block: false, alt: "1331")[$1 #hide($000$) 3 #hide($000$) 3 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 4")[$n = 4$]], [], [#math.equation(block: false, alt: "14641")[$1 #hide($000$) 4 #hide($000$) 6 #hide($000$) 4 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 5")[$n = 5$]], [], [#math.equation(block: false, alt: "15101051")[$1 #hide($000$) 5 #hide($i i 0$) 10 #hide($00$) 10 #hide($0 i i$) 5 #hide($000$) 1$]], [#math.equation(block: false, alt: "n equals 6")[$n = 6$]], [], [#math.equation(block: false, alt: "1615201561")[$1 #hide($000$) 6 #hide($i i 0$) 15 #hide($00$) 20 #hide($0 i i$) 15 #hide($000$) 6 #hide($i i 0$) 1$]], [#math.equation(block: false, alt: "n equals 7")[$n = 7$]], [], [#math.equation(block: false, alt: "172135352171")[$1 #hide($000$) 7 #hide($i i 0$) 21 #hide($00$) 35 #hide($0 i i$) 35 #hide($000$) 21 #hide($00$) 7 #hide($i i 0$) 1$]], [#math.equation(block: false, alt: "n equals 8")[$n = 8$]], [], [#math.equation(block: false, alt: "18285670562881")[$1 #hide($000$) 8 #hide($i i 0$) 28 #hide($00$) 56 #hide($0 i i$) 70 #hide($i i 0$) 56 #hide($00$) 28 #hide($000$) 8 #hide($i i 0$) 1$]], [#math.equation(block: false, alt: "n equals 9")[$n = 9$]], [], [#math.equation(block: false, alt: "193684126126843691")[$1 #hide($000$) 9 #hide($i i 0$) 36 #hide($00$) 84 #hide($0 i i$) 126 #hide($000$) 126 #hide($000$) 84 #hide($00$) 36 #hide($0 i i$) 9 #hide($i i 0$) 1$]], [#math.equation(block: false, alt: "n equals 10")[$n = 10$]], [], [#math.equation(block: false, alt: "1104512021025221012045101")[$1 #hide($000$) 10 #hide($i i 0$) 45 #hide($00$) 120 #hide($0 i i$) 210 #hide($000$) 252 #hide($i i 0$) 210 #hide($i i 0$) 120 #hide($i i 0$) 45 #hide($00$) 10 #hide($0 i i$) 1$]], )) Write down the portion of Pascal's triangle corresponding to rows from #math.equation(block: false, alt: "n equals 0")[$n = 0$] to #math.equation(block: false, alt: "n equals 12")[$n = 12$]. How many rows are involved? For Problems 7–18, write the power in expanded form. #math.equation(block: true, alt: "open parenthesis x plus 3 close parenthesis to the power 5")[$( x + 3 )^(5)$] #math.equation(block: true, alt: "x to the power 5 plus 15 x to the power 4 plus 90 x cubed plus 270 x squared plus 405 x plus 243")[$x^(5) + 15 x^(4) + 90 x^(3) + 270 x^(2) + 405 x + 243$] #math.equation(block: true, alt: "open parenthesis 2 x plus y close parenthesis to the power 4")[$( 2 x + y )^(4)$] #math.equation(block: true, alt: "open parenthesis z minus 3 close parenthesis to the power 4")[$( z − 3 )^(4)$] #math.equation(block: true, alt: "z to the power 4 minus 12 z cubed plus 54 z squared minus 108 z plus 81")[$z^(4) − 12 z^(3) + 54 z^(2) − 108 z + 81$] #math.equation(block: true, alt: "open parenthesis 2 w minus 1 close parenthesis to the power 5")[$( 2 w − 1 )^(5)$] #math.equation(block: true, alt: "open parenthesis 2 x minus the fraction y over 2 close parenthesis cubed")[$attach(( 2 x − display(frac(y, 2)) ), t: 3)$] #math.equation(block: true, alt: "8 x cubed minus 6 x squared y plus the fraction 3 over 2 x y squared minus the fraction 1 over 8 y cubed")[$8 x^(3) − 6 x^(2) y + display(frac(3, 2)) x y^(2) − display(frac(1, 8)) y^(3)$] #math.equation(block: true, alt: "open parenthesis the fraction x over 3 plus 3 close parenthesis to the power 6")[$attach(( display(frac(x, 3)) + 3 ), t: 6)$] #math.equation(block: true, alt: "open parenthesis x squared minus 3 close parenthesis to the power 7")[$( x^(2) − 3 )^(7)$] #math.equation(block: true, alt: "x to the power 14 minus 21 x to the power 12 plus 189 x to the power 10 minus 945 x to the power 8 plus 2835 x to the power 6 minus 5103 x to the power 4 plus 5103 x squared minus 2187")[$x^(14) − 21 x^(12) + 189 x^(10) − 945 x^(8) + 2835 x^(6) − 5103 x^(4) + 5103 x^(2) − 2187$] #math.equation(block: true, alt: "open parenthesis 1 minus y squared close parenthesis to the power 5")[$attach(( 1 − y^(2) ), t: 5)$] #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis to the power 5")[$( x + y )^(5)$] #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 plus 5 x y to the power 4 plus y to the power 5")[$x^(5) + 5 x^(4) y + 10 x^(3) y^(2) + 10 x^(2) y^(3) + 5 x y^(4) + y^(5)$] #math.equation(block: true, alt: "open parenthesis x minus y close parenthesis to the power 6")[$( x − y )^(6)$] #math.equation(block: true, alt: "open parenthesis p minus 2 q close parenthesis to the power 4")[$( p − 2 q )^(4)$] #math.equation(block: true, alt: "p to the power 4 minus 8 p cubed q plus 24 p squared q squared minus 32 p q cubed minus 16 p to the power 4")[$p^(4) − 8 p^(3) q + 24 p^(2) q^(2) − 32 p q^(3) − 16 p^(4)$] #math.equation(block: true, alt: "open parenthesis m plus 3 n close parenthesis to the power 8")[$( m + 3 n )^(8)$] Simplify #math.equation(block: false, alt: "open parenthesis 1 plus 5 t close parenthesis cubed plus open parenthesis 1 minus 5 t close parenthesis cubed")[$( 1 + 5 t )^(3) + ( 1 − 5 t )^(3)$]. #math.equation(block: true, alt: "2 plus 150 t squared")[$2 + 150 t^(2)$] Simplify #math.equation(block: false, alt: "open parenthesis 3 plus 2 a close parenthesis to the power 4 plus open parenthesis 3 minus 2 a close parenthesis to the power 4.")[$( 3 + 2 a )^(4) + ( 3 − 2 a )^(4) .$] Write #math.equation(block: false, alt: "open parenthesis z minus the fraction 1 over z close parenthesis to the power 5")[$attach(( z − display(frac(1, z)) ), t: 5)$] in descending powers of #math.equation(block: false, alt: "z")[$z$]. #math.equation(block: true, alt: "z to the power 5 minus 5 z cubed plus 10 z minus 10 z to the power minus 1 plus 5 z to the power minus 3 minus z to the power minus 5")[$z^(5) − 5 z^(3) + 10 z − 10 z^(− 1) + 5 z^(− 3) − z^(− 5)$] Write #math.equation(block: false, alt: "open parenthesis v plus the fraction 1 over v close parenthesis to the power 4")[$attach(( v + display(frac(1, v)) ), t: 4)$] in descending powers of #math.equation(block: false, alt: "v")[$v$]. Write the expressions in Problems 23–24 in expanded form and simplify. + #math.equation(block: false, alt: "5 !")[$5 !$] + #math.equation(block: false, alt: "the fraction 9 ! over 7 !")[$display(frac(9 !, 7 !))$] + #math.equation(block: false, alt: "the fraction 5 ! 7 ! over 12 !")[$display(frac(5 ! " " 7 !, 12 !))$] + #math.equation(block: false, alt: "the fraction 8 ! over 2 ! open parenthesis 8 minus 2 close parenthesis !")[$display(frac(8 !, 2 ! ( 8 − 2 ) !))$] + #math.equation(block: false, alt: "5 times 4 times 3 times 2 times 1 equals 120")[$5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120$] + #math.equation(block: false, alt: "the fraction 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 over 7 times 6 times 5 times 4 times 3 times 2 times 1 equals 72")[$display(frac(9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1, 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)) = 72$] + #math.equation(block: false, alt: "the fraction open parenthesis 5 times 4 times 3 times 2 times 1 close parenthesis open parenthesis 7 times 6 times 5 times 4 times 3 times 2 times 1 close parenthesis over 12 times 11 times 10 times 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 equals the fraction 1 over 792")[$display(frac(( 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ) ( 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ), 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)) = display(frac(1, 792))$] + #math.equation(block: false, alt: "the fraction 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 over open parenthesis 2 times 1 close parenthesis open parenthesis 6 times 5 times 4 times 3 times 2 times 1 close parenthesis equals 28")[$display(frac(8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1, ( 2 ⋅ 1 ) ( 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ))) = 28$] + #math.equation(block: false, alt: "7 !")[$7 !$] + #math.equation(block: false, alt: "the fraction 12 ! over 11 !")[$display(frac(12 !, 11 !))$] + #math.equation(block: false, alt: "the fraction 12 ! 4 ! over 16 !")[$display(frac(12 ! " " 4 !, 16 !))$] + #math.equation(block: false, alt: "the fraction 10 ! over 4 ! open parenthesis 10 minus 4 close parenthesis !")[$display(frac(10 !, 4 ! ( 10 − 4 ) !))$] In Problems 25–26, evaluate the binomial coefficient. + #math.equation(block: false, alt: "9 C sub 6")[$attach(" ", b: 9) C_(6) " "$] + #math.equation(block: false, alt: "12 C sub 3")[$attach(" ", b: 12) C_(3) " "$] + #math.equation(block: false, alt: "20 C sub 18")[$attach(" ", b: 20) C_(18) " "$] + #math.equation(block: false, alt: "14 C sub 9")[$attach(" ", b: 14) C_(9) " "$] + #math.equation(block: false, alt: "84")[$84$] + #math.equation(block: false, alt: "220")[$220$] + #math.equation(block: false, alt: "190")[$190$] + #math.equation(block: false, alt: "2002")[$2002$] + #math.equation(block: false, alt: "8 C sub 5")[$attach(" ", b: 8) C_(5) " "$] + #math.equation(block: false, alt: "13 C sub 4")[$attach(" ", b: 13) C_(4) " "$] + #math.equation(block: false, alt: "18 C sub 16")[$attach(" ", b: 18) C_(16) " "$] + #math.equation(block: false, alt: "16 C sub 7")[$attach(" ", b: 16) C_(7) " "$] Find in ascending powers of #math.equation(block: false, alt: "x")[$x$] the first three terms in the expansions of + #math.equation(block: false, alt: "open parenthesis 1 minus 2 x close parenthesis to the power 7")[$( 1 − 2 x )^(7)$] + #math.equation(block: false, alt: "open parenthesis 2 minus x close parenthesis to the power 6")[$( 2 − x )^(6)$] + #math.equation(block: false, alt: "1 minus 14 x plus 84 x squared")[$1 − 14 x + 84 x^(2)$] + #math.equation(block: false, alt: "64 minus 192 x plus 240 x squared")[$64 − 192 x + 240 x^(2)$] Find in descending powers of #math.equation(block: false, alt: "y")[$y$] the first three terms in the expansions of + #math.equation(block: false, alt: "open parenthesis 5 y minus 2 close parenthesis to the power 4")[$( 5 y − 2 )^(4)$] + #math.equation(block: false, alt: "open parenthesis 3 y plus 1 close parenthesis to the power 5")[$( 3 y + 1 )^(5)$] Find in descending powers of #math.equation(block: false, alt: "u")[$u$] the first three non-zero terms of #math.equation(block: false, alt: "open parenthesis 4 u minus 1 close parenthesis to the power 5 minus open parenthesis 4 u plus 1 close parenthesis to the power 5")[$( 4 u − 1 )^(5) − ( 4 u + 1 )^(5)$]. #math.equation(block: true, alt: "minus 2560 u to the power 4 minus 320 u squared minus 2")[$− 2560 u^(4) − 320 u^(2) − 2$] Find in ascending powers of #math.equation(block: false, alt: "t")[$t$] the first three non-zero terms of #math.equation(block: false, alt: "open parenthesis 1 minus 3 t close parenthesis to the power 6 minus open parenthesis 1 plus 3 t close parenthesis to the power 6")[$( 1 − 3 t )^(6) − ( 1 + 3 t )^(6)$]. Find the first four terms of + #math.equation(block: false, alt: "open parenthesis 1 minus 5 c close parenthesis to the power 6")[$( 1 − 5 c )^(6)$] + #math.equation(block: false, alt: "open parenthesis 1 minus 4 c close parenthesis open parenthesis 1 minus 5 c close parenthesis to the power 6")[$( 1 − 4 c ) ( 1 − 5 c )^(6)$] + #math.equation(block: false, alt: "1 minus 30 c plus 375 c squared minus 2500 c cubed")[$1 − 30 c + 375 c^(2) − 2500 c^(3)$] + #math.equation(block: false, alt: "1 minus 34 c plus 495 c squared minus 4000 c cubed")[$1 − 34 c + 495 c^(2) − 4000 c^(3)$] Find the first four terms of + #math.equation(block: false, alt: "open parenthesis 1 plus 2 a close parenthesis to the power 9")[$( 1 + 2 a )^(9)$] + #math.equation(block: false, alt: "open parenthesis 1 plus 3 a close parenthesis open parenthesis 1 plus 2 a close parenthesis to the power 9")[$( 1 + 3 a ) ( 1 + 2 a )^(9)$] Simplify #math.equation(block: false, alt: "open parenthesis 1 plus the square root of 3 close parenthesis to the power 4 plus open parenthesis 1 minus the square root of 3 close parenthesis to the power 4")[$attach(( 1 + sqrt(3) ), t: 4) + attach(( 1 − sqrt(3) ), t: 4)$]. #math.equation(block: true, alt: "56")[$56$] Simplify #math.equation(block: false, alt: "open parenthesis the square root of 2 plus the square root of 3 close parenthesis to the power 4 plus open parenthesis the square root of 2 minus the square root of 3 close parenthesis to the power 4")[$attach(( sqrt(2) + sqrt(3) ), t: 4) + attach(( sqrt(2) − sqrt(3) ), t: 4)$]. For Problems 35–42, find the coefficient of the indicated term. #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis to the power 20 ; x to the power 13 y to the power 7")[$( x + y )^(20) ; " " " " x^(13) y^(7)$] #math.equation(block: true, alt: "77 , 520")[$77 , 520$] #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis to the power 15 ; x to the power 12 y cubed")[$( x + y )^(15) ; " " " " x^(12) y^(3)$] #math.equation(block: true, alt: "open parenthesis a minus 2 b close parenthesis to the power 12 ; a to the power 5 b to the power 7")[$( a − 2 b )^(12) ; " " " " a^(5) b^(7)$] #math.equation(block: true, alt: "minus 101 , 376")[$− 101 , 376$] #math.equation(block: true, alt: "open parenthesis 2 a minus b close parenthesis to the power 12 ; a to the power 8 b to the power 4")[$( 2 a − b )^(12) ; " " " " a^(8) b^(4)$] #math.equation(block: true, alt: "open parenthesis x minus the square root of 2 close parenthesis to the power 10 ; x to the power 4")[$attach(( x − sqrt(2) ), t: 10) ; " " " " x^(4)$] #math.equation(block: true, alt: "1680")[$1680$] #math.equation(block: true, alt: "open parenthesis x minus the fraction 1 over 2 close parenthesis to the power 8 ; x to the power 5")[$attach(( x − display(frac(1, 2)) ), t: 8) ; " " " " x^(5)$] #math.equation(block: true, alt: "open parenthesis a cubed minus b cubed close parenthesis to the power 9 ; a to the power 18 b to the power 9")[$attach(( a^(3) − b^(3) ), t: 9) ; " " " " a^(18) b^(9)$] #math.equation(block: true, alt: "minus 84")[$− 84$] #math.equation(block: true, alt: "open parenthesis x squared minus y squared close parenthesis to the power 7 ; x to the power 10 y to the power 4")[$attach(( x^(2) − y^(2) ), t: 7) ; " " " " x^(10) y^(4)$] For Problems 43–44, refer to Pascal's triangle to answer the questions. Write out the terms of the sequence #math.equation(block: false, alt: "11 to the power n")[$11^(n)$] for #math.equation(block: false, alt: "n equals 0 , 1 , 2 , 3 , 4")[$n = 0 , 1 , 2 , 3 , 4$]. How are these related to the rows of Pascal's triangle? Can you explain why? (Hint: Rewrite #math.equation(block: false, alt: "11")[$11$] as #math.equation(block: false, alt: "10 plus 1")[$10 + 1$].) #math.equation(block: false, alt: "1 , 11 , 121 , 1331 , 14641.")[$1 , " " 11 , " " 121 , " " 1331 , " " 14641. " " " "$] The digits of the terms in the sequence correspond to the numbers in the first five rows of Pascal's triangle. If #math.equation(block: false, alt: "11 to the power n equals open parenthesis 10 plus 1 close parenthesis to the power n")[$11^(n) = ( 10 + 1 )^(n)$] is expanded as a binomial, each term is the product of a number from Pascal's triangle times a power of 10 times a power of 1. Write out the terms of the sequence #math.equation(block: false, alt: "1.1 to the power n")[$1.1^(n)$] for #math.equation(block: false, alt: "n equals 0 , 1 , 2 , 3 , 4")[$n = 0 , 1 , 2 , 3 , 4$]. How are these related to the rows of Pascal's triangle? Can you explain why? (Hint: Rewrite #math.equation(block: false, alt: "1.1")[$1.1$] as #math.equation(block: false, alt: "1 plus 0.1")[$1 + 0.1$].)