#set document(title: "9.1 Sequences", 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")) == 9.1#h(0.6em)Sequences === Definitions and Notation Consider the following function: Gwynn would like to compete in a triathlon, but she needs to improve her swimming. She begins a training schedule in which she swims 20 laps a day for the first week and increases that number by 6 laps each week. The table below gives the first few values of the function. #figure(table( columns: 10, align: left, inset: 6pt, table.header([Week], [#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")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "n")[$n$]]), [Number of Laps], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "26")[$26$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "38")[$38$]], [#math.equation(block: false, alt: "44")[$44$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "56")[$56$]], [#math.equation(block: false, alt: "62")[$62$]], [#math.equation(block: false, alt: "f open parenthesis n close parenthesis")[$f ( n )$]], )) The function makes sense only for input values that are positive integers. We would not ask how many laps Gwynn swims in week 4.63, or in week #math.equation(block: false, alt: "minus 6")[$− 6$]. A function whose inputs are a set of successive positive integers is called a #strong[sequence]. Most people think of a sequence as a list of objects in which the order is important, and we often present a mathematical sequence in just that way. The information in the table above can be displayed more simply by listing only the output values, in order: #math.equation(block: true, alt: "20 , 26 , 32 , 38 , 44 , 50 , 56 , 62")[$20 , " " " " 26 , " " " " 32 , " " " " 38 , " " " " 44 , " " " " 50 , " " " " 56 , " " " " 62$] When we list them in this way, the output values are called the #strong[terms] of the sequence. The input values are indicated implicitly by the position of the term. For example, the third term of the sequence, 32, is the value of #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis")[$f ( 3 )$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ We often use the notation #math.equation(block: false, alt: "a sub n")[$a_(n)$] instead of #math.equation(block: false, alt: "f open parenthesis n close parenthesis")[$f ( n )$] to refer to the terms of a sequence. Thus, #math.equation(block: false, alt: "a sub 1 equals f open parenthesis 1 close parenthesis , a sub 2 equals f open parenthesis 2 close parenthesis ,")[$a_(1) = f ( 1 ) , " " " " a_(2) = f ( 2 ) ,$] and so on. ] #examplebox("Example 1")[][ Let #math.equation(block: false, alt: "a sub n")[$a_(n)$] be the number of seats in the #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$] row of a theater. Write an equation using subscript notation for each of the following statements. + The first row has 30 seats. + The twentieth row has 68 seats. + Row 18 has twice as many seats as row 2. + Row #math.equation(block: false, alt: "n")[$n$] has #math.equation(block: false, alt: "2 n plus 28")[$2 n + 28$] seats. + Row #math.equation(block: false, alt: "n plus 1")[$n + 1$] has 2 more seats than the previous row. #solutionbox[ + #math.equation(block: false, alt: "a sub 1")[$a_(1)$] represents the number of seats in row 1, so #math.equation(block: false, alt: "a sub 1 equals 30")[$a_(1) = 30$]. + #math.equation(block: false, alt: "a sub 20")[$a_(20)$] represents the number of seats in row 20, so #math.equation(block: false, alt: "a sub 20 equals 68")[$a_(20) = 68$]. + #math.equation(block: false, alt: "a sub 18 equals 2 a sub 2")[$a_(18) = 2 a_(2)$] + #math.equation(block: false, alt: "a sub n equals 2 n plus 28")[$a_(n) = 2 n + 28$] + #math.equation(block: false, alt: "a sub n plus 1 equals 2 plus a sub n")[$a_(n + 1) = 2 + a_(n)$]. (Row #math.equation(block: false, alt: "n")[$n$] is the row before row #math.equation(block: false, alt: "n plus 1")[$n + 1$].) ] ] In Example 1, #math.equation(block: false, alt: "a sub n equals 2 n plus 28")[$a_(n) = 2 n + 28$] is a formula in terms of #math.equation(block: false, alt: "n")[$n$] for the number of seats in the #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$] row of the theater. The expression #math.equation(block: false, alt: "2 n plus 28")[$2 n + 28$] is called the #strong[general term] for the sequence. Let #math.equation(block: false, alt: "A sub n")[$A_(n)$] be the number of dollars in a bank account at the end of its #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$]year. Write an equation using subscript notation for each of the following statements. + The account had \$1149.90 at the end of the second year. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ \[Note: Enter "A1" to get #math.equation(block: false, alt: "A sub 1")[$A_(1)$], etc.\] + At the end of year 4 the account had \$85.75 more than at the end of year 3. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ + At the end of year 5 the account had 1.07 times as much as at the end of year 4. #linebreak() \_\_\_\_\_#math.equation(block: false, alt: "equals")[$=$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "A sub 2 equals 1149.90")[$A_(2) = 1149.90$] + #math.equation(block: false, alt: "A sub 4 equals 85.75 plus A sub 3")[$A_(4) = 85.75 + A_(3)$] + #math.equation(block: false, alt: "A sub 5 equals 1.07 A sub 4")[$A_(5) = 1.07 A_(4)$] ] Let #math.equation(block: false, alt: "A sub n")[$A_(n)$] be the number of dollars in a bank account at the end of its #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$]year. Write an equation using subscript notation for each of the following statements. + The account had \$1149.90 at the end of the second year. + At the end of year 4 the account had \$85.75 more than at the end of year 3. + At the end of year 5 the account had 1.07 times as much as at the end of year 4. #solutionbox[ + #math.equation(block: false, alt: "A sub 2 equals 1149.90")[$A_(2) = 1149.90$] + #math.equation(block: false, alt: "A sub 4 equals 85.75 plus A sub 3")[$A_(4) = 85.75 + A_(3)$] + #math.equation(block: false, alt: "A sub 5 equals 1.07 A sub 4")[$A_(5) = 1.07 A_(4)$] ] #examplebox("Example 2")[][ Find the first four terms in each sequence with the given general term. + #math.equation(block: false, alt: "a sub n equals the fraction n open parenthesis n plus 1 close parenthesis over 2")[$a_(n) = display(frac(n ( n + 1 ), 2))$] + #math.equation(block: false, alt: "a sub n equals open parenthesis minus 1 close parenthesis to the power n 2 to the power n")[$a_(n) = ( − 1 )^(n) 2^(n)$] #solutionbox[ We evaluate each general term for successive values of #math.equation(block: false, alt: "n")[$n$]. + - #math.equation(block: false, alt: "a sub 1 equals the fraction 1 open parenthesis 1 plus 1 close parenthesis over 2 equals 1")[$a_(1) = display(frac(1 ( 1 + 1 ), 2)) = 1$] - #math.equation(block: false, alt: "a sub 2 equals the fraction 2 open parenthesis 2 plus 1 close parenthesis over 2 equals 3")[$a_(2) = display(frac(2 ( 2 + 1 ), 2)) = 3$] - #math.equation(block: false, alt: "a sub 3 equals the fraction 3 open parenthesis 3 plus 1 close parenthesis over 2 equals 6")[$a_(3) = display(frac(3 ( 3 + 1 ), 2)) = 6$] - #math.equation(block: false, alt: "a sub 4 equals the fraction 4 open parenthesis 4 plus 1 close parenthesis over 2 equals 10")[$a_(4) = display(frac(4 ( 4 + 1 ), 2)) = 10$] #linebreak() The first four terms are 1, 3, 6, and 10. + - #math.equation(block: false, alt: "a sub 1 equals open parenthesis minus 1 close parenthesis to the power 12 to the power 1 equals minus 2")[$a_(1) = ( − 1 )^(1) 2^(1) = − 2$] - #math.equation(block: false, alt: "a sub 2 equals open parenthesis minus 1 close parenthesis squared 2 squared equals 4")[$a_(2) = ( − 1 )^(2) 2^(2) = 4$] - #math.equation(block: false, alt: "a sub 3 equals open parenthesis minus 1 close parenthesis cubed 2 cubed equals minus 8")[$a_(3) = ( − 1 )^(3) 2^(3) = − 8$] - #math.equation(block: false, alt: "a sub 4 equals open parenthesis minus 1 close parenthesis to the power 42 to the power 4 equals 16")[$a_(4) = ( − 1 )^(4) 2^(4) = 16$] #linebreak() The first four terms are #math.equation(block: false, alt: "minus 2 ,")[$− 2 ,$] #math.equation(block: false, alt: "4 ,")[$4 ,$] #math.equation(block: false, alt: "minus 8 ,")[$− 8 ,$] and #math.equation(block: false, alt: "16")[$16$]. ] ] How is a sequence different from an ordinary function? \_\_\_\_\_ #solutionbox[ answer ] How is a sequence different from an ordinary function? + Its outputs are integers. + Its inputs are integers. + There is a formula for its outputs. + The outputs must be positive.. Find the first four terms in the sequence with the general term #math.equation(block: false, alt: "b sub n equals open parenthesis n plus 1 close parenthesis squared minus n squared")[$b_(n) = ( n + 1 )^(2) − n^(2)$]. Answer: \_\_\_\_\_ (List the terms in order; separate terms with a comma. #solutionbox[ 3, 5, 7, 9 ] Find the first four terms in the sequence with the general term #math.equation(block: false, alt: "b sub n equals open parenthesis n plus 1 close parenthesis squared minus n squared")[$b_(n) = ( n + 1 )^(2) − n^(2)$]. #solutionbox[ 3, 5, 7, 9 ] === Applications of Sequences Sequences are useful for describing situations that have discrete values, rather than continuous values. For some savings accounts, interest is posted only once a year. #examplebox("Example 3")[][ You deposit \$8000 in a savings account that pays 5% interest compounded annually. How much money will be in the account at the end of each of the next 4 years? #solutionbox[ During the first year the account will earn 5% of \$8000, or #math.equation(block: false, alt: "0.05 open parenthesis 8000 close parenthesis equals 400")[$0.05 ( 8000 ) = 400$] dollars. Thus, at the end of the first year the account will contain the original \$8000 plus the \$400 interest for a total of \$8400. At the end of the second year the account will have the \$8400 from the previous year, plus 5% of \$8400, or #math.equation(block: false, alt: "0.05 open parenthesis 8400 close parenthesis equals 420")[$0.05 ( 8400 ) = 420$] dollars in interest. We can write this sum as #math.equation(block: true, alt: "8400 plus 8400 open parenthesis 0.05 close parenthesis equals 8400 open parenthesis 1 plus 0.05 close parenthesis equals 8400 open parenthesis 1.05 close parenthesis dollars.")[$8400 + 8400 ( 0.05 ) = 8400 ( 1 + 0.05 ) = 8400 ( 1.05 ) " " " " "dollars."$] Thus, at the end of the second year there will be #math.equation(block: false, alt: "$ 8400 open parenthesis 1.05 close parenthesis equals $ 8820")[$\$ 8400 ( 1.05 ) = \$ 8820$] in the account. In fact, each new balance is found by multiplying the previous balance by 1.05. The balances at the ends of the third and fourth years are #math.equation(block: true, alt: "$ 8820 open parenthesis 1.05 close parenthesis equals $ 9261 and $ 9261 open parenthesis 1.05 close parenthesis equals $ 9724.05")[$upright(\$) 8820 ( 1.05 ) = upright(\$) 9261 " " " " " " " " "and" " " " " " " " " upright(\$) 9261 ( 1.05 ) = upright(\$) 9724.05$] The annual balances form the sequence 8400, 8820, 9261, and 9724.05, as shown below. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "a sub n")[$a_(n)$]]), [1], [8400], [2], [8820], [3], [9261], [4], [9724.05], )) ] ] You just finished your last cup of coffee and have 100 mg of caffeine in your system. For each hour that passes, the amount of caffeine in your system decreases by 14%. How much caffeine is in your system at the start of each of the next 4 hours? Answer: \_\_\_\_\_ mg (List the terms in order; separate terms with a comma. #solutionbox[ 100 mg, 86 mg, 74 mg, 63.6 mg approximately ] You just finished your last cup of coffee and have 100 mg of caffeine in your system. For each hour that passes, the amount of caffeine in your system decreases by 14%. How much caffeine is in your system at the start of each of the next 4 hours? #solutionbox[ 100 mg, 86 mg, 74 mg, 63.6 mg approximately ] The terms of a sequence \_\_\_\_\_ #solutionbox[ answer ] The terms of a sequence + describe its formula. + are its boundary values. + are its outputs. + are either increasing or decreasing. #examplebox("Example 4")[][ You deposit \$8000 in a savings account that pays 5% simple annual interest. (The interest is earned only on the initial \$8000 and not compounded.) How much money will be in the account at the end of each of the next 4 years? #solutionbox[ The account earns 5% of \$8000 or #math.equation(block: false, alt: "0.05 open parenthesis 8000 close parenthesis equals 400")[$0.05 ( 8000 ) = 400$] dollars during the first year. At the end of the first year the account will contain the original \$8000 plus \$400 interest for a total of \$8400. During the second year interest will be earned only on the initial \$8000 deposited, for another 400 dollars. So at the end of the second year the account will have the \$8400 from the previous year plus \$400 interest, for a total of #math.equation(block: true, alt: "8400 plus 400 equals 8800 dollars")[$8400 + 400 = 8800 " " "dollars"$] In fact, the balance at the end of each succeeding year is found by adding \$400 simple interest to the previous balance. The annual balances form the sequence 8400, 8800, 9200, and 9600, as shown below. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "b sub n")[$b_(n)$]]), [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "8400")[$8400$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "8800")[$8800$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "9200")[$9200$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "9600")[$9600$]], )) ] ] The financing agreement on your \$12,000 car requires you to pay \$1200 now and \$208.87 a month for 5 years. What is the total amount you have paid after each of the first 4 payments? Answer: \$\_\_\_\_\_ (List the terms in order; separate terms with a comma. #solutionbox[ \$1200, \$1408.87, \$1617.74, \$1826.61 ] The financing agreement on your \$12,000 car requires you to pay \$1200 now and \$208.87 a month for 5 years. What is the total amount you have paid after each of the first 4 payments? #solutionbox[ \$1200, \$1408.87, \$1617.74, \$1826.61 ] #examplebox("Example 5")[][ A regular polygon is a geometric figure in which all the sides are equal in length. For example, an equilateral triangle and a square are regular polygons. All the interior angles in a regular polygon are equal also. Find the general term #math.equation(block: false, alt: "a sub n")[$a_(n)$] for the sequence that gives the size of an interior angle in a regular polygon of #math.equation(block: false, alt: "n")[$n$] sides. #figure(figph[regular polygons], alt: "regular polygons", caption: none) #solutionbox[ This sequence starts with #math.equation(block: false, alt: "a sub 3")[$a_(3)$] because we cannot have a polygon with fewer than three sides. We already know the values of #math.equation(block: false, alt: "a sub 3")[$a_(3)$] and #math.equation(block: false, alt: "a sub 4")[$a_(4)$]: each angle in an equilateral triangle is #math.equation(block: false, alt: "60 degrees")[$60^(∘)$], and each angle in a square is #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. We would like to find a formula for the size of the angles in any regular polygon. If we can find the sum of all the angles in a regular polygon, we can divide by #math.equation(block: false, alt: "n")[$n$] to find the size of each. (You can check that this idea works for the equilateral triangle and the square.) To find the sum of the angles, notice that any polygon can be partitioned into triangles, as shown below. #figure(figph[polygon divided into triangles], alt: "polygon divided into triangles", caption: none) By sketching some examples, convince yourself that every polygon of #math.equation(block: false, alt: "n")[$n$] sides (for #math.equation(block: false, alt: "n greater than or equal to 3")[$n ≥ 3$]) can be partitioned into #math.equation(block: false, alt: "n minus 2")[$n − 2$] triangles. Because the angles in every triangle add up to #math.equation(block: false, alt: "180 degrees")[$180^(∘)$], the sum of the angles in an #math.equation(block: false, alt: "n")[$n$]-sided polygon is #math.equation(block: false, alt: "open parenthesis n minus 2 close parenthesis")[$( n − 2 )$] times #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. To find the size of just one of the angles, we divide the sum by #math.equation(block: false, alt: "n")[$n$]. This gives us the general term of the sequence: #math.equation(block: true, alt: "a sub n equals the fraction open parenthesis n minus 2 close parenthesis 180 over n")[$a_(n) = display(frac(( n − 2 ) 180, n))$] ] ] === Recursively Defined Sequences A sequence is defined #strong[recursively] if each term of the sequence is defined in terms of its predecessors. For example, the sequence defined by #math.equation(block: true, alt: "a sub 1 equals 2 , a sub n plus 1 equals 3 a sub n minus 2")[$a_(1) = 2 , " " " " " " " " a_(n + 1) = 3 a_(n) − 2$] is a recursive sequence. Its first four terms are #math.equation(block: true, alt: "a sub 1, equals 2; a sub 2, equals 3 a sub 1 minus 2 equals 3 open parenthesis 2 close parenthesis minus 2 equals 4; a sub 3, equals 3 a sub 2 minus 2 equals 3 open parenthesis 4 close parenthesis minus 2 equals 10; a sub 4, equals 3 a sub 3 minus 2 equals 3 open parenthesis 10 close parenthesis minus 2 equals 28")[$a_(1) & = 2 \ a_(2) & = 3 a_(1) − 2 = 3 ( 2 ) − 2 = 4 \ a_(3) & = 3 a_(2) − 2 = 3 ( 4 ) − 2 = 10 \ a_(4) & = 3 a_(3) − 2 = 3 ( 10 ) − 2 = 28$] #examplebox("Example 6")[][ Make a table showing the first five terms of the recursive sequence #math.equation(block: true, alt: "a sub 1 equals minus 1 , a sub n plus 1 equals open parenthesis a sub n close parenthesis squared minus 4")[$a_(1) = − 1 , " " " " " " " " " " " " a_(n + 1) = ( a_(n) )^(2) − 4$] #solutionbox[ The first term is given. We find each new term by using the recursive formula: #math.equation(block: true, alt: "a sub 2, equals open parenthesis a sub 1 close parenthesis squared minus 4 equals open parenthesis minus 1 close parenthesis squared minus 4 equals minus 3; a sub 3, equals open parenthesis a sub 2 close parenthesis squared minus 4 equals open parenthesis minus 3 close parenthesis squared minus 4 equals 5; a sub 4, equals open parenthesis a sub 3 close parenthesis squared minus 4 equals 5 squared minus 4 equals 21; a sub 5, equals open parenthesis a sub 4 close parenthesis squared minus 4 equals 21 squared minus 4 equals 437")[$a_(2) & = ( a_(1) )^(2) − 4 = ( − 1 )^(2) − 4 = − 3 \ a_(3) & = ( a_(2) )^(2) − 4 = ( − 3 )^(2) − 4 = 5 \ a_(4) & = ( a_(3) )^(2) − 4 = 5^(2) − 4 = 21 \ a_(5) & = ( a_(4) )^(2) − 4 = 21^(2) − 4 = 437$] The first five terms are 1, 3, 5, 21, and 437, as shown in the table. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#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")[$5$]]), [#math.equation(block: false, alt: "a sub n")[$a_(n)$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "21")[$21$]], [#math.equation(block: false, alt: "437")[$437$]], )) ] ] The definition of a recursive sequence must include a starting point, usually the first term of the sequence, and a formula for calculating the next term of the sequence in terms of the previous term (or terms). #examplebox("Example 7")[][ You deposit \$8000 in a savings account that pays 5% interest compounded annually. Find a recursive definition for the sequence of account balances if the money is kept in the account for #math.equation(block: false, alt: "n")[$n$] years. #solutionbox[ If we use the letter #math.equation(block: false, alt: "a")[$a$] for that sequence, we have #math.equation(block: false, alt: "a sub 1 equals 8400")[$a_(1) = 8400$]. Each successive term was obtained by multiplying the previous term by 1.05, which means that #math.equation(block: true, alt: "a sub n equals 1.05 a sub n minus 1")[$a_(n) = 1.05 a_(n − 1)$] The sequence is determined by the recursive definition #math.equation(block: true, alt: "a sub 1 equals 8400 , a sub n equals 1.05 a sub n minus 1")[$a_(1) = 8400 , " " " " " " " " a_(n) = 1.05 a_(n − 1)$] ] ] Find a recursive definition for the sequence in CheckPoint 9.8 above if you do not drink any more coffee for #math.equation(block: false, alt: "n")[$n$] hours. #math.equation(block: false, alt: "C sub 1 equals")[$C_(1) =$]\_\_\_\_\_, #math.equation(block: false, alt: "C sub n plus 1 equals")[$C_(n + 1) =$]\_\_\_\_\_ \[Note: Enter "Cn" to get #math.equation(block: false, alt: "C sub n")[$C_(n)$].\] #solutionbox[ #math.equation(block: true, alt: "C sub 1 equals 100")[$C_(1) = 100$], #math.equation(block: true, alt: "C sub n plus 1 equals 0.86 C sub n")[$C_(n + 1) = 0.86 C_(n)$] ] Find a recursive definition for the sequence in CheckPoint 9.8 above if you do not drink any more coffee for #math.equation(block: false, alt: "n")[$n$] hours. #solutionbox[ #math.equation(block: true, alt: "C sub 1 equals 100")[$C_(1) = 100$], #math.equation(block: true, alt: "C sub n plus 1 equals 0.86 C sub n")[$C_(n + 1) = 0.86 C_(n)$] ] #examplebox("Example 8")[][ You deposit \$8000 in a savings account that pays 5% simple annual interest. Find a recursive definition for the sequence of account balances if the money is kept in the account for #math.equation(block: false, alt: "n")[$n$] years. #solutionbox[ If we use the letter #math.equation(block: false, alt: "b")[$b$] for that sequence, we have #math.equation(block: false, alt: "b sub 1 equals 8400")[$b_(1) = 8400$]. Each successive term was obtained by adding 400 to the previous term, which means that #math.equation(block: true, alt: "b sub n equals b sub n minus 1 plus 400")[$b_(n) = b_(n − 1) + 400$] The sequence is determined by the recursive definition #math.equation(block: true, alt: "b sub 1 equals 8400 , b sub n equals b sub n minus 1 plus 400")[$b_(1) = 8400 , " " " " " " b_(n) = b_(n − 1) + 400$] ] ] If each output of a sequence is defined in terms of earlier outputs, we say the sequence is \_\_\_\_\_ #solutionbox[ answer ] If each output of a sequence is defined in terms of earlier outputs, we say the sequence is + general. + discrete. + ordered. + recursive. #examplebox("Example 9")[][ Karen joins a savings plan in which she deposits \$200 per month and receives 12% annual interest compounded monthly. + Find a recursively defined sequence that gives the amount of money in Karen's account #math.equation(block: false, alt: "n")[$n$] months later. + Find the first four terms of the sequence. #solutionbox[ + In the first month Karen deposits \$200, so #math.equation(block: false, alt: "a sub 1 equals 200")[$a_(1) = 200$]. Each month thereafter Karen receives 1% interest (one-twelfth of 12% annual interest) on the previous month's balance, and then adds \$200 to the total. For example, before she makes her deposit in the second month the account has #math.equation(block: true, alt: "200 plus 0.01 open parenthesis 200 close parenthesis equals 1.01 open parenthesis 200 close parenthesis dollars")[$200 + 0.01 ( 200 ) = 1.01 ( 200 ) " " "dollars"$] She then adds \$200 to this amount for a total of #math.equation(block: true, alt: "a sub 2 equals 1.01 open parenthesis 200 close parenthesis plus 200 dollars")[$a_(2) = 1.01 ( 200 ) + 200 " " "dollars"$] In general, after the #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$] deposit Karen's account contains #math.equation(block: false, alt: "a sub n")[$a_(n)$] dollars. In the next month she earns 1% interest on that balance, giving her #math.equation(block: true, alt: "a sub n plus 1.01 a sub n equals 1.01 a sub n dollars")[$a_(n) + 1.01 a_(n) = 1.01 a_(n) " " "dollars"$] Then she deposits another \$200 for a total of #math.equation(block: true, alt: "a sub n plus 1 equals 1.01 a sub n plus 200 dollars")[$a_(n + 1) = 1.01 a_(n) + 200 " " "dollars"$] Thus, the recursive sequence is defined by #math.equation(block: true, alt: "a sub 1 equals 200 , a sub n plus 1 equals 1.01 a sub n plus 200 dollars")[$a_(1) = 200 , " " " " " " a_(n + 1) = 1.01 a_(n) + 200 " " "dollars"$] + We evaluate the general formula found in part (a) for #math.equation(block: false, alt: "n equals 1 , 2 , 3 , 4 :")[$n = 1 , 2 , 3 , 4 :$] #math.equation(block: true, alt: "a sub 1, equals 200; a sub 2, equals 1.01 a sub 1 plus 200; equals 1.01 open parenthesis 200 close parenthesis plus 200 equals 402; a sub 3, equals 1.01 a sub 2 plus 200; equals 1.01 open parenthesis 402 close parenthesis plus 200 equals 606.02; a sub 4, equals 1.01 a sub 3 plus 200; equals 1.01 open parenthesis 606.02 close parenthesis plus 200 equals 812.08")[$a_(1) & = 200 \ a_(2) & = 1.01 a_(1) + 200 \ & = 1.01 ( 200 ) + 200 = 402 \ a_(3) & = 1.01 a_(2) + 200 \ & = 1.01 ( 402 ) + 200 = 606.02 \ a_(4) & = 1.01 a_(3) + 200 \ & = 1.01 ( 606.02 ) + 200 = 812.08$] ] ] === Section Summary ==== Vocabulary - Sequence - General term - Recursive sequence ==== CONCEPTS + A function whose inputs are a set of successive positive integers is called a #strong[sequence]. + The output values are called the #strong[terms] of the sequence. + A formula in terms of #math.equation(block: false, alt: "n")[$n$] for the #math.equation(block: false, alt: "n to the power t h")[$n^(t h)$] term of a sequence is called the #strong[general term] of the sequence. + A sequence is defined #strong[recursively] if each term of the sequence is defined in terms of its predecessors. ==== STUDY QUESTIONS + What distinguishes a sequence from an ordinary function? + What are the range values of a sequence called? + Give an example of a situation in which a sequence is more appropriate than a function whose domain is an interval of real numbers. + What is a recursively defined sequence? How is such a sequence defined? ==== SKILLS Practice each skill in the Homework Problems listed. + Evaluate the general term \#1–14, \#21–26, + Use subscript notation \#15–20 + Evaluate a recursively defined sequence \#27–34, \#45–52 + Write a formula for a sequence \#35–44 === Homework 9.1 For Problems 1–14, find the first four terms in the sequence whose general term is given. #math.equation(block: true, alt: "a sub n equals n minus 5")[$a_(n) = n − 5$] #math.equation(block: true, alt: "minus 4 , minus 3 , minus 2 , minus 1")[$− 4 , − 3 , − 2 , − 1$] #math.equation(block: true, alt: "b sub n equals 2 n minus 3")[$b_(n) = 2 n − 3$] #math.equation(block: true, alt: "c sub n equals the fraction n squared minus 2 over 2")[$c_(n) = display(frac(n^(2) − 2, 2))$] #math.equation(block: true, alt: "the fraction minus 1 over 2 , 1 , the fraction 7 over 2 , 7")[$display(frac(− 1, 2)) , 1 , display(frac(7, 2)) , 7$] #math.equation(block: true, alt: "d sub n equals the fraction 3 over n squared plus 1")[$d_(n) = display(frac(3, n^(2) + 1))$] #math.equation(block: true, alt: "s sub n equals 1 plus the fraction 1 over n")[$s_(n) = 1 + display(frac(1, n))$] #math.equation(block: true, alt: "2 , 1.5 , 1.3 bar , 1.25")[$2 , 1.5 , 1. limits(3)^(―) , 1.25$] #math.equation(block: true, alt: "t sub n equals the fraction n over 2 n minus 1")[$t_(n) = display(frac(n, 2 n − 1))$] #math.equation(block: true, alt: "u sub n equals the fraction n open parenthesis n minus 1 close parenthesis over 2")[$u_(n) = display(frac(n ( n − 1 ), 2))$] #math.equation(block: true, alt: "0 , 1 , 3 , 6")[$0 , 1 , 3 , 6$] #math.equation(block: true, alt: "v sub n equals the fraction 5 over n open parenthesis n plus 1 close parenthesis")[$v_(n) = display(frac(5, n ( n + 1 )))$] #math.equation(block: true, alt: "w sub n equals open parenthesis minus 1 close parenthesis to the power n")[$w_(n) = ( − 1 )^(n)$] #math.equation(block: true, alt: "minus 1 , 1 , minus 1 , 1")[$− 1 , 1 , − 1 , 1$] #math.equation(block: true, alt: "A sub n equals open parenthesis minus 1 close parenthesis to the power n plus 1")[$A_(n) = ( − 1 )^(n + 1)$] #math.equation(block: true, alt: "B sub n equals the fraction open parenthesis minus 1 close parenthesis to the power n open parenthesis n minus 2 close parenthesis over n")[$B_(n) = display(frac(( − 1 )^(n) ( n − 2 ), n))$] #math.equation(block: true, alt: "1 , 0 , the fraction minus 1 over 3 , the fraction 1 over 2")[$1 , 0 , display(frac(− 1, 3)) , display(frac(1, 2))$] #math.equation(block: true, alt: "C sub n equals open parenthesis minus 1 close parenthesis to the power n minus 13 to the power n plus 1")[$C_(n) = ( − 1 )^(n − 1) 3^(n + 1)$] #math.equation(block: true, alt: "D sub n equals 1")[$D_(n) = 1$] #math.equation(block: true, alt: "1 , 1 , 1 , 1")[$1 , 1 , 1 , 1$] #math.equation(block: true, alt: "E sub n equals minus 1")[$E_(n) = − 1$] For Problems 15–20, suppose that #math.equation(block: false, alt: "a sub 1 , a sub 2 , a sub 3 , . . .")[$a_(1) , a_(2) , a_(3) , . . .$] is a sequence. Write an equation using subscript notation for each sentence. The first term of the sequence is #math.equation(block: false, alt: "the fraction 4 over 3")[$display(frac(4, 3))$]. #math.equation(block: true, alt: "a sub 1 equals the fraction 4 over 3")[$a_(1) = display(frac(4, 3))$] The third term of the sequence is 8 more than the second term. The #math.equation(block: false, alt: "n to the power th")[$n^("th")$] term is 3 times the previous term. #math.equation(block: true, alt: "a sub n equals 3 a sub n minus 1")[$a_(n) = 3 a_(n − 1)$] The #math.equation(block: false, alt: "n to the power th")[$n^("th")$] term is one third of the term that follows it. The #math.equation(block: false, alt: "open parenthesis n plus 1 close parenthesis to the power st")[$( n + 1 )^("st")$] term is one third of the term that follows it. #math.equation(block: true, alt: "a sub n plus 1 equals the fraction 1 over 3 a sub n plus 2")[$a_(n + 1) = display(frac(1, 3)) a_(n + 2)$] The #math.equation(block: false, alt: "open parenthesis n plus 1 close parenthesis to the power st")[$( n + 1 )^("st")$] term is 3 times the previous term. For Problems 21–26, find the indicated term for each sequence. #math.equation(block: true, alt: "D sub n equals 2 to the power n minus n ; D sub 6")[$D_(n) = 2^(n) − n ; " " " " " " D_(6)$] #math.equation(block: true, alt: "58")[$58$] #math.equation(block: true, alt: "E sub n equals the square root of n plus 1 ; E sub 11")[$E_(n) = sqrt(n + 1) ; " " " " " " E_(11)$] #math.equation(block: true, alt: "x sub n equals log ⁡ n ; x sub 26")[$x_(n) = log n ; " " " " " " x_(26)$] #math.equation(block: true, alt: "1.415")[$1.415$] #math.equation(block: true, alt: "y sub n equals log ⁡ open parenthesis n plus 1 close parenthesis ; y sub 9")[$y_(n) = log ( n + 1 ) ; " " " " " " y_(9)$] #math.equation(block: true, alt: "z sub n equals 2 the square root of n ; z sub 20")[$z_(n) = 2 sqrt(n) ; " " " " " " z_(20)$] #math.equation(block: true, alt: "8.944")[$8.944$] #math.equation(block: true, alt: "U sub n equals the fraction n plus 1 over n minus 1 ; U sub 17")[$U_(n) = display(frac(n + 1, n − 1)) ; " " " " " " U_(17)$] For Problems 27-34, make a table showing the first five terms of the recursively defined sequence. #math.equation(block: true, alt: "s sub 1 equals 3 ; s sub n equals s sub n minus 1 plus 2")[$s_(1) = 3 ; " " " " " " s_(n) = s_(n − 1) + 2$] #math.equation(block: true, alt: "3 , 5 , 7 , 9 , 11")[$3 , 5 , 7 , 9 , 11$] #math.equation(block: true, alt: "c sub 1 equals 6 ; c sub n equals c sub n minus 1 minus 4")[$c_(1) = 6 ; " " " " " " c_(n) = c_(n − 1) − 4$] #math.equation(block: true, alt: "d sub 1 equals 24 ; d sub n plus 1 equals the fraction minus 1 over 2 d sub n")[$d_(1) = 24 ; " " " " " " d_(n + 1) = display(frac(− 1, 2)) d_(n)$] #math.equation(block: true, alt: "24 , minus 12 , 6 , minus 3 , 1.5")[$24 , − 12 , 6 , − 3 , 1.5$] #math.equation(block: true, alt: "r sub 1 equals 27 ; r sub n plus 1 equals the fraction 2 over 3 r sub n")[$r_(1) = 27 ; " " " " " " r_(n + 1) = display(frac(2, 3)) r_(n)$] #math.equation(block: true, alt: "t sub 1 equals 1 ; t sub n plus 1 equals open parenthesis n plus 1 close parenthesis t sub n")[$t_(1) = 1 ; " " " " " " t_(n + 1) = ( n + 1 ) t_(n)$] #math.equation(block: true, alt: "1 , 2 , 6 , 24 , 120")[$1 , 2 , 6 , 24 , 120$] #math.equation(block: true, alt: "x sub 1 equals 1 ; x sub n plus 1 equals open parenthesis the fraction n over n minus 1 close parenthesis x sub n")[$x_(1) = 1 ; " " " " " " x_(n + 1) = ( display(frac(n, n − 1)) ) x_(n)$] #math.equation(block: true, alt: "w sub 1 equals 100 ; w sub n equals 1.10 w sub n minus 1 plus 100")[$w_(1) = 100 ; " " " " " " w_(n) = 1.10 w_(n − 1) + 100$] #math.equation(block: true, alt: "100 , 210 , 331 , 464.1 , 610.51")[$100 , 210 , 331 , 464.1 , 610.51$] #math.equation(block: true, alt: "q sub 1 equals 100 ; q sub n equals 0.9 q sub n minus 1 plus 100")[$q_(1) = 100 ; " " " " " " q_(n) = 0.9 q_(n − 1) + 100$] For Problems 35–42, + Make a table showing the first four terms of each sequence. + Write an equation to define the sequence recursively. A new car costs \$14,000 and depreciates in value by 15% each year. How much is the car worth after #math.equation(block: false, alt: "n")[$n$] years? + #math.equation(block: false, alt: "14 , 000 ; 11 , 900 ; 10 , 115 ; 8597.75")[$14 , 000 ; 11 , 900 ; 10 , 115 ; 8597.75$] + #math.equation(block: false, alt: "v sub 1 equals 14 , 000 ; v sub n plus 1 equals 0.85 v sub n")[$v_(1) = 14 , 000 ; v_(n + 1) = 0.85 v_(n)$] Krishna takes a job as an executive secretary for \$21,000 per year with a guaranteed 5% raise each year. What will his salary be after #math.equation(block: false, alt: "n")[$n$] years? A long distance phone call costs \$1.10 to make the connection and an additional \$0.45 for each minute. What is the cost of a call that lasts #math.equation(block: false, alt: "n")[$n$] minutes? + #math.equation(block: false, alt: "1.55 , 2.00 , 2.45 , 2.90")[$1.55 , 2.00 , 2.45 , 2.90$] + #math.equation(block: false, alt: "c sub 1 equals 1.55 ; c sub n plus 1 equals c sub n plus 0.45")[$c_(1) = 1.55 ; c_(n + 1) = c_(n) + 0.45$] Bettina earns \$1000 per month plus \$57 for each satellite dish that she sells. What is her monthly income when she sells #math.equation(block: false, alt: "n")[$n$] satellite dishes? Geraldo inherits an annuity of \$50,000 that earns 12% annual interest compounded monthly. If he withdraws \$500 at the end of each month, what is the value of the annuity after #math.equation(block: false, alt: "n")[$n$] months? + #math.equation(block: false, alt: "50 , 000 ; 50 , 000 ; 50 , 000 ; 50 , 000")[$50 , 000 ; 50 , 000 ; 50 , 000 ; 50 , 000$] + #math.equation(block: false, alt: "v sub 1 equals v sub n")[$v_(1) = v_(n)$] Eve borrowed \$18,000 for a new car at 6% annual interest compounded monthly. If she pays \$400 per month toward the loan, how much does she owe after #math.equation(block: false, alt: "n")[$n$] months? Majel must take 10 milliliters of a medication directly into her bloodstream at constant intervals. During each time interval her kidneys filter out 20% of the drug present just after the most recent dose. How much of the drug will be in her bloodstream after #math.equation(block: false, alt: "n")[$n$] doses? + #math.equation(block: false, alt: "10 , 18 , 24.4 , 29.52")[$10 , 18 , 24.4 , 29.52$] + #math.equation(block: false, alt: "d sub 1 equals 10 ; d sub n plus 1 equals 0.8 d sub n plus 10")[$d_(1) = 10 ; d_(n + 1) = 0.8 d_(n) + 10$] A forest contains 64,000 trees. According to a new logging plan, each year 5% of the trees will be cut down and 16,000 new trees will be planted. How many trees will be in the forest after #math.equation(block: false, alt: "n")[$n$] years? + Draw three non-collinear points in the plane. (The points should not lie on the same line.) How many distinct lines are determined by the points? (In other words, how many different lines can you draw by choosing two of the points and joining them?) + Add a fourth point to your diagram. Now how many lines are determined? + Let #math.equation(block: false, alt: "L sub n")[$L_(n)$] stand for the number of distinct lines determined by non-collinear points. Make a table showing the first five terms of the sequence. + Find a recursive formula for the sequence #math.equation(block: false, alt: "L sub n")[$L_(n)$]. + 3 + 6 + 0, 1, 3, 6, 10 + #math.equation(block: false, alt: "L sub 1 equals 0 , L sub n plus 1 equals n plus L sub n")[$L_(1) = 0 , L_(n + 1) = n + L_(n)$] + Draw two distinct non-parallel lines in the plane. In how many points do the lines intersect? + Add a third line to your diagram that is not parallel to either of the first two lines. How many intersection points are there? + Let #math.equation(block: false, alt: "P sub n")[$P_(n)$] stand for the number of intersection points determined by lines in the plane, no two of which are parallel. Make a table showing the first five terms of the sequence. + Find a recursive formula for the sequence #math.equation(block: false, alt: "P sub n")[$P_(n)$]. The Fibonacci sequence is found throughout nature. For example, the numbers of spirals in a sunflower or on a pineapple are elements of the Fibonacci sequence. It is named after the Italian mathematician Fibonnaci, who used it to model the growth of a population of rabbits. The Fibonacci sequence is defined recursively by #math.equation(block: true, alt: "f sub 1 equals 1 , f sub 2 equals 1 , f sub n plus 2 equals f sub n plus f sub n plus 1")[$f_(1) = 1 , " " " " f_(2) = 1 , " " " " f_(n + 2) = f_(n) + f_(n + 1)$] + Make a table showing the first 16 terms of the Fibonacci sequence. + Calculate the quotients #math.equation(block: false, alt: "the fraction f sub n plus 1 over f sub n")[$display(frac(f_(n + 1), f_(n)))$] for #math.equation(block: false, alt: "n equals 1")[$n = 1$] to #math.equation(block: false, alt: "n equals 15")[$n = 15$]. What do you observe? Now find a decimal approximation for the golden ratio, #math.equation(block: false, alt: "the fraction 1 plus the square root of 5 over 2")[$display(frac(1 + sqrt(5), 2))$]. + #math.equation(block: false, alt: "1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987")[$1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , 610 , 987$] + #math.equation(block: false, alt: "1 , 2 , 1.5 , 1.6 bar , 1.6 , 1.625 , 1.615 , 1.619 , 1.618 , 1.618 , 1.618 , 1.618 , 1.618 ,")[$1 , 2 , 1.5 , 1. limits(6)^(―) , 1.6 , 1.625 , 1.615 , 1.619 , 1.618 , 1.618 , 1.618 , 1.618 , 1.618 ,$] #linebreak() #math.equation(block: false, alt: "1.618 , 1.618.")[$1.618 , 1.618 .$] #linebreak() The quotients approach a limit near #math.equation(block: false, alt: "1.618")[$1.618$]. #linebreak() #math.equation(block: false, alt: "the fraction 1 plus the square root of 5 over 2 approximately equals 1.618033989")[$display(frac(1 + sqrt(5), 2)) ≈ 1.618033989$], the same as the limit above. The Lucas sequence is defined recursively by #math.equation(block: true, alt: "L sub 1 equals 2 , L sub 2 equals 1 , L sub n plus 2 equals L sub n plus L sub n plus 1")[$L_(1) = 2 , " " " " L_(2) = 1 , " " " " L_(n + 2) = L_(n) + L_(n + 1)$] + Find the first 10 terms of the Lucas sequence. + Calculate #math.equation(block: false, alt: "open parenthesis L sub n close parenthesis squared minus L sub n open parenthesis L sub n plus 2 close parenthesis")[$( L_(n) )^(2) − L_(n) ( L_(n + 2) )$] for #math.equation(block: false, alt: "n equals 1")[$n = 1$] to #math.equation(block: false, alt: "n equals 8")[$n = 8$]. What do you notice? For Problems 47–52, use a calculator to evaluate a large number of terms for each recursive sequence. What happens to the terms as #math.equation(block: false, alt: "n")[$n$] gets larger? #math.equation(block: true, alt: "a sub 1 equals 1 ; a sub n equals the fraction 1 over 1 plus a sub n minus 1 plus 1")[$a_(1) = 1 ; " " " " " " a_(n) = display(frac(1, 1 + a_(n − 1))) + 1$] #math.equation(block: false, alt: "a sub n")[$a_(n)$] approaches #math.equation(block: false, alt: "1.4142")[$1.4142$], or #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$]. #math.equation(block: true, alt: "b sub 1 equals 1 ; b sub n equals the fraction 2 over 1 plus b sub n minus 1 plus 1")[$b_(1) = 1 ; " " " " " " b_(n) = display(frac(2, 1 + b_(n − 1))) + 1$] #math.equation(block: true, alt: "c sub 1 equals 3 ; c sub n equals the fraction the square root of 1 plus c sub n minus 1 over 2")[$c_(1) = 3 ; " " " " " " c_(n) = display(frac(sqrt(1 + c_(n − 1)), 2))$] #math.equation(block: false, alt: "c sub n")[$c_(n)$] approaches #math.equation(block: false, alt: "0.64039")[$0.64039$], or #math.equation(block: false, alt: "the fraction 1 plus the square root of 17 over 8")[$display(frac(1 + sqrt(17), 8))$] #math.equation(block: true, alt: "d sub 1 equals 8 ; d sub n equals the fraction the square root of 1 plus d sub n minus 1 over 2")[$d_(1) = 8 ; " " " " " " d_(n) = display(frac(sqrt(1 + d_(n − 1)), 2))$] #math.equation(block: true, alt: "s sub 1 equals 1 ; s sub n equals the fraction 1 over 2 open parenthesis s sub n minus 1 plus the fraction 4 over s sub n minus 1 close parenthesis")[$s_(1) = 1 ; " " " " " " s_(n) = display(frac(1, 2)) ( s_(n − 1) + display(frac(4, s_(n − 1))) )$] #math.equation(block: false, alt: "s sub n")[$s_(n)$] approaches #math.equation(block: false, alt: "2")[$2$] #math.equation(block: true, alt: "t sub 1 equals 1 ; t sub n equals the fraction 1 over 2 open parenthesis t sub n minus 1 plus the fraction 9 over t sub n minus 1 close parenthesis")[$t_(1) = 1 ; " " " " " " t_(n) = display(frac(1, 2)) ( t_(n − 1) + display(frac(9, t_(n − 1))) )$]