{"id":1692,"date":"2021-07-23T16:43:09","date_gmt":"2021-07-23T16:43:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1692"},"modified":"2022-03-21T23:00:55","modified_gmt":"2022-03-21T23:00:55","slug":"terminology-of-sequences","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/terminology-of-sequences\/","title":{"raw":"Terminology of Sequences","rendered":"Terminology of Sequences"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the formula for the general term of a sequence<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1169739025144\" data-depth=\"1\">\r\n<p id=\"fs-id1169739202521\">To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form<\/p>\r\n\r\n<div id=\"fs-id1169739062458\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{1},{a}_{2},{a}_{3}\\text{,}\\ldots,{a}_{n}\\text{,}\\ldots\\text{.}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738826814\">Each of the numbers in the sequence is called a term. The symbol [latex]n[\/latex] is called the index variable for the sequence. We use the notation<\/p>\r\n\r\n<div id=\"fs-id1169739376228\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\left\\{{a}_{n}\\right\\}}_{n=1}^{\\infty },\\text{or simply}\\left\\{{a}_{n}\\right\\}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739188531\">to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number [latex]{a}_{n}[\/latex] exists for each positive integer [latex]n[\/latex], we can also define a sequence as a function whose domain is the set of positive integers.<\/p>\r\n<p id=\"fs-id1169736855866\">Let\u2019s consider the infinite, ordered list<\/p>\r\n\r\n<div id=\"fs-id1169738936091\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2,4,8,16,32\\text{,}\\ldots[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739202470\">This is a sequence in which the first, second, and third terms are given by [latex]{a}_{1}=2[\/latex], [latex]{a}_{2}=4[\/latex], and [latex]{a}_{3}=8[\/latex]. You can probably see that the terms in this sequence have the following pattern:<\/p>\r\n\r\n<div id=\"fs-id1169738849565\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{1}={2}^{1},{a}_{2}={2}^{2},{a}_{3}={2}^{3},{a}_{4}={2}^{4},\\text{and }{a}_{5}={2}^{5}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736601138\">Assuming this pattern continues, we can write the [latex]n\\text{th}[\/latex] term in the sequence by the explicit formula [latex]{a}_{n}={2}^{n}[\/latex]. Using this notation, we can write this sequence as<\/p>\r\n\r\n<div id=\"fs-id1169739307977\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\left\\{{2}^{n}\\right\\}}_{n=1}^{\\infty }\\text{or}\\left\\{{2}^{n}\\right\\}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739282664\">Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the [latex]n\\text{th}[\/latex] term [latex]{a}_{n}[\/latex] in terms of the previous term [latex]{a}_{n - 1}[\/latex]. In particular, we can define this sequence as the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] where [latex]{a}_{1}=2[\/latex] and for all [latex]n\\ge 2[\/latex], each term [latex]{a}_{n}[\/latex] is defined by the <span data-type=\"term\">recurrence relation<\/span> [latex]{a}_{n}=2{a}_{n - 1}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169738878532\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169738888670\">An <strong>infinite sequence<\/strong> [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is an ordered list of numbers of the form<\/p>\r\n\r\n<div id=\"fs-id1169739188460\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{1},{a}_{2}\\text{,}\\ldots,{a}_{n}\\text{,}\\ldots\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-id1169738962610\">The subscript [latex]n[\/latex] is called the <strong>index variable<\/strong> of the sequence. Each number [latex]{a}_{n}[\/latex] is a <span data-type=\"term\">term<\/span> of the sequence. Sometimes sequences are defined by <strong>explicit formulas<\/strong>, in which case [latex]{a}_{n}=f\\left(n\\right)[\/latex] for some function [latex]f\\left(n\\right)[\/latex] defined over the positive integers. In other cases, sequences are defined by using a <strong>recurrence relation<\/strong>. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169738826555\">Note that the index does not have to start at [latex]n=1[\/latex] but could start with other integers. For example, a sequence given by the explicit formula [latex]{a}_{n}=f\\left(n\\right)[\/latex] could start at [latex]n=0[\/latex], in which case the sequence would be<\/p>\r\n\r\n<div id=\"fs-id1169739038407\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{0},{a}_{1},{a}_{2}\\text{,}\\ldots\\text{.}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738829461\">Similarly, for a sequence defined by a recurrence relation, the term [latex]{a}_{0}[\/latex] may be given explicitly, and the terms [latex]{a}_{n}[\/latex] for [latex]n\\ge 1[\/latex] may be defined in terms of [latex]{a}_{n - 1}[\/latex]. Since a sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] has exactly one value for each positive integer [latex]n[\/latex], it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] consists of all points [latex]\\left(n,{a}_{n}\\right)[\/latex] for all positive integers [latex]n[\/latex]. Figure 1 shows the graph of [latex]\\left\\{{2}^{n}\\right\\}[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_01_001\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234240\/CNX_Calc_Figure_09_01_001.jpg\" alt=\"A graph in quadrant one containing the following points: (1, 2), (2, 4), (3, 8), (4, 16).\" width=\"487\" height=\"388\" data-media-type=\"image\/jpeg\" \/> Figure 1. The plotted points are a graph of the sequence [latex]\\left\\{{2}^{n}\\right\\}[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1169739111058\">Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an <strong>arithmetic sequence<\/strong>, the <em data-effect=\"italics\">difference<\/em> between every pair of consecutive terms is the same. For example, consider the sequence<\/p>\r\n\r\n<div id=\"fs-id1169738824405\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3,7,11,15,19\\text{,}\\ldots\\text{.}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738883815\">You can see that the difference between every consecutive pair of terms is [latex]4[\/latex]. Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation<\/p>\r\n\r\n<div id=\"fs-id1169739274427\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\bigg\\{ \\begin{array}{c} {a}_{1}=3\\hfill \\\\ {a}_{n}={a}_{n-1}+4 \\text{ for } n\\ge 2 \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739199490\">Note that<\/p>\r\n\r\n<div id=\"fs-id1169738891016\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{a}_{2}=3+4\\\\ {a}_{3}=3+4+4=3+2\\cdot 4\\\\ {a}_{4}=3+4+4+4=3+3\\cdot 4.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738982334\">Thus the sequence can also be described using the explicit formula<\/p>\r\n\r\n<div id=\"fs-id1169739000061\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {a}_{n}&amp; =3+4\\left(n - 1\\right)\\hfill \\\\ &amp; =4n - 1.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739194788\">In general, an arithmetic sequence is any sequence of the form [latex]{a}_{n}=cn+b[\/latex].<\/p>\r\n<p id=\"fs-id1169739037240\">In a <strong>geometric sequence<\/strong>, the <em data-effect=\"italics\">ratio<\/em> of every pair of consecutive terms is the same. For example, consider the sequence<\/p>\r\n\r\n<div id=\"fs-id1169739096795\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2,-\\frac{2}{3},\\frac{2}{9},-\\frac{2}{27},\\frac{2}{81}\\text{,}\\ldots\\text{.}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739105061\">We see that the ratio of any term to the preceding term is [latex]-\\frac{1}{3}[\/latex]. Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as<\/p>\r\n\r\n<div id=\"fs-id1169736655039\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ {a}_{1}=2\\hfill \\\\ {a}_{n}=-\\frac{1}{3}\\cdot {a}_{n - 1}\\text{ for }n\\ge 2.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736791594\">Alternatively, since<\/p>\r\n\r\n<div id=\"fs-id1169739024768\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {a}_{2}=-\\frac{1}{3}\\cdot 2\\hfill \\\\ {a}_{3}=\\left(-\\frac{1}{3}\\right)\\left(-\\frac{1}{3}\\right)\\left(2\\right)={\\left(-\\frac{1}{3}\\right)}^{2}\\cdot 2\\hfill \\\\ {a}_{4}=\\left(-\\frac{1}{3}\\right)\\left(-\\frac{1}{3}\\right)\\left(-\\frac{1}{3}\\right)\\left(2\\right)={\\left(-\\frac{1}{3}\\right)}^{3}\\cdot 2,\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738924575\">we see that the sequence can be described by using the explicit formula<\/p>\r\n\r\n<div id=\"fs-id1169739109807\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}=2{\\left(-\\frac{1}{3}\\right)}^{n - 1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739102205\">The sequence [latex]\\left\\{{2}^{n}\\right\\}[\/latex] that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is [latex]2[\/latex]. In general, a geometric sequence is any sequence of the form [latex]{a}_{n}=c{r}^{n}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169736659986\" data-type=\"example\">\r\n<div id=\"fs-id1169739234198\" data-type=\"exercise\">\r\n<div id=\"fs-id1169739098012\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding Explicit Formulas<\/h3>\r\n<div id=\"fs-id1169739098012\" data-type=\"problem\">\r\n<p id=\"fs-id1169739302356\">For each of the following sequences, find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence.<\/p>\r\n\r\n<ol id=\"fs-id1169736611796\" type=\"a\">\r\n \t<li>[latex]-\\frac{1}{2},\\frac{2}{3},-\\frac{3}{4},\\frac{4}{5},-\\frac{5}{6}\\text{,}\\ldots[\/latex]<\/li>\r\n \t<li>[latex]\\frac{3}{4},\\frac{9}{7},\\frac{27}{10},\\frac{81}{13},\\frac{243}{16}\\text{,}\\ldots[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1169738915710\" data-type=\"solution\">\r\n<ol id=\"fs-id1169738941818\" type=\"a\">\r\n \t<li>First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are negative, and the even terms are positive. Therefore, the [latex]n\\text{th}[\/latex] term includes a factor of [latex]{\\left(-1\\right)}^{n}[\/latex]. Next, consider the sequence of numerators [latex]\\left\\{1,2,3\\text{,}\\ldots\\right\\}[\/latex] and the sequence of denominators [latex]\\left\\{2,3,4\\text{,}\\ldots\\right\\}[\/latex]. We can see that both of these sequences are arithmetic sequences. The [latex]n\\text{th}[\/latex] term in the sequence of numerators is [latex]n[\/latex], and the [latex]n\\text{th}[\/latex] term in the sequence of denominators is [latex]n+1[\/latex]. Therefore, the sequence can be described by the explicit formula<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738993914\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n}n}{n+1}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>The sequence of numerators [latex]3,9,27,81,243\\text{,}\\ldots[\/latex] is a geometric sequence. The numerator of the [latex]n\\text{th}[\/latex] term is [latex]{3}^{n}[\/latex] The sequence of denominators [latex]4,7,10,13,16\\text{,}\\ldots[\/latex] is an arithmetic sequence. The denominator of the [latex]n\\text{th}[\/latex] term is [latex]4+3\\left(n - 1\\right)=3n+1[\/latex]. Therefore, we can describe the sequence by the explicit formula [latex]{a}_{n}=\\frac{{3}^{n}}{3n+1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739016124\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169739223015\" data-type=\"exercise\">\r\n<div id=\"fs-id1169736593628\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169736593628\" data-type=\"problem\">\r\n<p id=\"fs-id1169739242722\">Find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence [latex]\\left\\{\\frac{1}{5},-\\frac{1}{7},\\frac{1}{9},-\\frac{1}{11}\\text{,}\\ldots\\right\\}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1169738880222\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739045640\">The denominators form an arithmetic sequence.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1169739093464\" data-type=\"solution\">\r\n<p id=\"fs-id1169738966727\">[latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n+1}}{3+2n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oduNoQPFDD4?controls=0&amp;start=285&amp;end=348&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.1.1_285to348_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.1\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1169739010866\" data-type=\"example\">\r\n<div id=\"fs-id1169738890687\" data-type=\"exercise\">\r\n<div id=\"fs-id1169739182834\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Defined by Recurrence Relations<\/h3>\r\n<div id=\"fs-id1169739182834\" data-type=\"problem\">\r\n<p id=\"fs-id1169736617641\">For each of the following recursively defined sequences, find an explicit formula for the sequence.<\/p>\r\n\r\n<ol id=\"fs-id1169739350728\" type=\"a\">\r\n \t<li>[latex]{a}_{1}=2[\/latex], [latex]{a}_{n}=-3{a}_{n - 1}[\/latex] for [latex]n\\ge 2[\/latex]<\/li>\r\n \t<li>[latex]{a}_{1}=\\frac{1}{2}[\/latex], [latex]{a}_{n}={a}_{n - 1}+{\\left(\\frac{1}{2}\\right)}^{n}[\/latex] for [latex]n\\ge 2[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1169739027829\" data-type=\"solution\">\r\n<ol id=\"fs-id1169738999234\" type=\"a\">\r\n \t<li>Writing out the first few terms, we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738860999\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{a}_{1}=2\\hfill \\\\ {a}_{2}=-3{a}_{1}=-3\\left(2\\right)\\hfill \\\\ {a}_{3}=-3{a}_{2}={\\left(-3\\right)}^{2}2\\hfill \\\\ {a}_{4}=-3{a}_{3}={\\left(-3\\right)}^{3}2.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIn general,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736709052\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}=2{\\left(-3\\right)}^{n - 1}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Write out the first few terms:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738999519\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {a}_{1}=\\frac{1}{2}\\hfill \\\\ {a}_{2}={a}_{1}+{\\left(\\frac{1}{2}\\right)}^{2}=\\frac{1}{2}+\\frac{1}{4}=\\frac{3}{4}\\hfill \\\\ {a}_{3}={a}_{2}+{\\left(\\frac{1}{2}\\right)}^{3}=\\frac{3}{4}+\\frac{1}{8}=\\frac{7}{8}\\hfill \\\\ {a}_{4}={a}_{3}+{\\left(\\frac{1}{2}\\right)}^{4}=\\frac{7}{8}+\\frac{1}{16}=\\frac{15}{16}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFrom this pattern, we derive the explicit formula<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739019569\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}=\\frac{{2}^{n}-1}{{2}^{n}}=1-\\frac{1}{{2}^{n}}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739016786\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169738936238\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738975437\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169738975437\" data-type=\"problem\">\r\n<p id=\"fs-id1169739214209\">Find an explicit formula for the sequence defined recursively such that [latex]{a}_{1}=-4[\/latex] and [latex]{a}_{n}={a}_{n - 1}+6[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1169739302833\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739102236\">This is an arithmetic sequence.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1169736854641\" data-type=\"solution\">\r\n<p id=\"fs-id1169739242787\">[latex]{a}_{n}=6n - 10[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oduNoQPFDD4?controls=0&amp;start=475&amp;end=506&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.1.1_475to506_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.1\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]19431[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1169738981819\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the formula for the general term of a sequence<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1169739025144\" data-depth=\"1\">\n<p id=\"fs-id1169739202521\">To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form<\/p>\n<div id=\"fs-id1169739062458\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{1},{a}_{2},{a}_{3}\\text{,}\\ldots,{a}_{n}\\text{,}\\ldots\\text{.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738826814\">Each of the numbers in the sequence is called a term. The symbol [latex]n[\/latex] is called the index variable for the sequence. We use the notation<\/p>\n<div id=\"fs-id1169739376228\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\left\\{{a}_{n}\\right\\}}_{n=1}^{\\infty },\\text{or simply}\\left\\{{a}_{n}\\right\\}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739188531\">to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number [latex]{a}_{n}[\/latex] exists for each positive integer [latex]n[\/latex], we can also define a sequence as a function whose domain is the set of positive integers.<\/p>\n<p id=\"fs-id1169736855866\">Let\u2019s consider the infinite, ordered list<\/p>\n<div id=\"fs-id1169738936091\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2,4,8,16,32\\text{,}\\ldots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739202470\">This is a sequence in which the first, second, and third terms are given by [latex]{a}_{1}=2[\/latex], [latex]{a}_{2}=4[\/latex], and [latex]{a}_{3}=8[\/latex]. You can probably see that the terms in this sequence have the following pattern:<\/p>\n<div id=\"fs-id1169738849565\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{1}={2}^{1},{a}_{2}={2}^{2},{a}_{3}={2}^{3},{a}_{4}={2}^{4},\\text{and }{a}_{5}={2}^{5}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736601138\">Assuming this pattern continues, we can write the [latex]n\\text{th}[\/latex] term in the sequence by the explicit formula [latex]{a}_{n}={2}^{n}[\/latex]. Using this notation, we can write this sequence as<\/p>\n<div id=\"fs-id1169739307977\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\left\\{{2}^{n}\\right\\}}_{n=1}^{\\infty }\\text{or}\\left\\{{2}^{n}\\right\\}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739282664\">Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the [latex]n\\text{th}[\/latex] term [latex]{a}_{n}[\/latex] in terms of the previous term [latex]{a}_{n - 1}[\/latex]. In particular, we can define this sequence as the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] where [latex]{a}_{1}=2[\/latex] and for all [latex]n\\ge 2[\/latex], each term [latex]{a}_{n}[\/latex] is defined by the <span data-type=\"term\">recurrence relation<\/span> [latex]{a}_{n}=2{a}_{n - 1}[\/latex].<\/p>\n<div id=\"fs-id1169738878532\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1169738888670\">An <strong>infinite sequence<\/strong> [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is an ordered list of numbers of the form<\/p>\n<div id=\"fs-id1169739188460\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{1},{a}_{2}\\text{,}\\ldots,{a}_{n}\\text{,}\\ldots\\text{.}[\/latex]<\/div>\n<p id=\"fs-id1169738962610\">The subscript [latex]n[\/latex] is called the <strong>index variable<\/strong> of the sequence. Each number [latex]{a}_{n}[\/latex] is a <span data-type=\"term\">term<\/span> of the sequence. Sometimes sequences are defined by <strong>explicit formulas<\/strong>, in which case [latex]{a}_{n}=f\\left(n\\right)[\/latex] for some function [latex]f\\left(n\\right)[\/latex] defined over the positive integers. In other cases, sequences are defined by using a <strong>recurrence relation<\/strong>. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169738826555\">Note that the index does not have to start at [latex]n=1[\/latex] but could start with other integers. For example, a sequence given by the explicit formula [latex]{a}_{n}=f\\left(n\\right)[\/latex] could start at [latex]n=0[\/latex], in which case the sequence would be<\/p>\n<div id=\"fs-id1169739038407\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{0},{a}_{1},{a}_{2}\\text{,}\\ldots\\text{.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738829461\">Similarly, for a sequence defined by a recurrence relation, the term [latex]{a}_{0}[\/latex] may be given explicitly, and the terms [latex]{a}_{n}[\/latex] for [latex]n\\ge 1[\/latex] may be defined in terms of [latex]{a}_{n - 1}[\/latex]. Since a sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] has exactly one value for each positive integer [latex]n[\/latex], it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] consists of all points [latex]\\left(n,{a}_{n}\\right)[\/latex] for all positive integers [latex]n[\/latex]. Figure 1 shows the graph of [latex]\\left\\{{2}^{n}\\right\\}[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_09_01_001\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234240\/CNX_Calc_Figure_09_01_001.jpg\" alt=\"A graph in quadrant one containing the following points: (1, 2), (2, 4), (3, 8), (4, 16).\" width=\"487\" height=\"388\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The plotted points are a graph of the sequence [latex]\\left\\{{2}^{n}\\right\\}[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1169739111058\">Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an <strong>arithmetic sequence<\/strong>, the <em data-effect=\"italics\">difference<\/em> between every pair of consecutive terms is the same. For example, consider the sequence<\/p>\n<div id=\"fs-id1169738824405\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3,7,11,15,19\\text{,}\\ldots\\text{.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738883815\">You can see that the difference between every consecutive pair of terms is [latex]4[\/latex]. Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation<\/p>\n<div id=\"fs-id1169739274427\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\bigg\\{ \\begin{array}{c} {a}_{1}=3\\hfill \\\\ {a}_{n}={a}_{n-1}+4 \\text{ for } n\\ge 2 \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739199490\">Note that<\/p>\n<div id=\"fs-id1169738891016\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{a}_{2}=3+4\\\\ {a}_{3}=3+4+4=3+2\\cdot 4\\\\ {a}_{4}=3+4+4+4=3+3\\cdot 4.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738982334\">Thus the sequence can also be described using the explicit formula<\/p>\n<div id=\"fs-id1169739000061\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {a}_{n}& =3+4\\left(n - 1\\right)\\hfill \\\\ & =4n - 1.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739194788\">In general, an arithmetic sequence is any sequence of the form [latex]{a}_{n}=cn+b[\/latex].<\/p>\n<p id=\"fs-id1169739037240\">In a <strong>geometric sequence<\/strong>, the <em data-effect=\"italics\">ratio<\/em> of every pair of consecutive terms is the same. For example, consider the sequence<\/p>\n<div id=\"fs-id1169739096795\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2,-\\frac{2}{3},\\frac{2}{9},-\\frac{2}{27},\\frac{2}{81}\\text{,}\\ldots\\text{.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739105061\">We see that the ratio of any term to the preceding term is [latex]-\\frac{1}{3}[\/latex]. Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as<\/p>\n<div id=\"fs-id1169736655039\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ {a}_{1}=2\\hfill \\\\ {a}_{n}=-\\frac{1}{3}\\cdot {a}_{n - 1}\\text{ for }n\\ge 2.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736791594\">Alternatively, since<\/p>\n<div id=\"fs-id1169739024768\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {a}_{2}=-\\frac{1}{3}\\cdot 2\\hfill \\\\ {a}_{3}=\\left(-\\frac{1}{3}\\right)\\left(-\\frac{1}{3}\\right)\\left(2\\right)={\\left(-\\frac{1}{3}\\right)}^{2}\\cdot 2\\hfill \\\\ {a}_{4}=\\left(-\\frac{1}{3}\\right)\\left(-\\frac{1}{3}\\right)\\left(-\\frac{1}{3}\\right)\\left(2\\right)={\\left(-\\frac{1}{3}\\right)}^{3}\\cdot 2,\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738924575\">we see that the sequence can be described by using the explicit formula<\/p>\n<div id=\"fs-id1169739109807\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}=2{\\left(-\\frac{1}{3}\\right)}^{n - 1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739102205\">The sequence [latex]\\left\\{{2}^{n}\\right\\}[\/latex] that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is [latex]2[\/latex]. In general, a geometric sequence is any sequence of the form [latex]{a}_{n}=c{r}^{n}[\/latex].<\/p>\n<div id=\"fs-id1169736659986\" data-type=\"example\">\n<div id=\"fs-id1169739234198\" data-type=\"exercise\">\n<div id=\"fs-id1169739098012\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding Explicit Formulas<\/h3>\n<div id=\"fs-id1169739098012\" data-type=\"problem\">\n<p id=\"fs-id1169739302356\">For each of the following sequences, find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence.<\/p>\n<ol id=\"fs-id1169736611796\" type=\"a\">\n<li>[latex]-\\frac{1}{2},\\frac{2}{3},-\\frac{3}{4},\\frac{4}{5},-\\frac{5}{6}\\text{,}\\ldots[\/latex]<\/li>\n<li>[latex]\\frac{3}{4},\\frac{9}{7},\\frac{27}{10},\\frac{81}{13},\\frac{243}{16}\\text{,}\\ldots[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738915710\" data-type=\"solution\">\n<ol id=\"fs-id1169738941818\" type=\"a\">\n<li>First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are negative, and the even terms are positive. Therefore, the [latex]n\\text{th}[\/latex] term includes a factor of [latex]{\\left(-1\\right)}^{n}[\/latex]. Next, consider the sequence of numerators [latex]\\left\\{1,2,3\\text{,}\\ldots\\right\\}[\/latex] and the sequence of denominators [latex]\\left\\{2,3,4\\text{,}\\ldots\\right\\}[\/latex]. We can see that both of these sequences are arithmetic sequences. The [latex]n\\text{th}[\/latex] term in the sequence of numerators is [latex]n[\/latex], and the [latex]n\\text{th}[\/latex] term in the sequence of denominators is [latex]n+1[\/latex]. Therefore, the sequence can be described by the explicit formula<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738993914\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n}n}{n+1}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>The sequence of numerators [latex]3,9,27,81,243\\text{,}\\ldots[\/latex] is a geometric sequence. The numerator of the [latex]n\\text{th}[\/latex] term is [latex]{3}^{n}[\/latex] The sequence of denominators [latex]4,7,10,13,16\\text{,}\\ldots[\/latex] is an arithmetic sequence. The denominator of the [latex]n\\text{th}[\/latex] term is [latex]4+3\\left(n - 1\\right)=3n+1[\/latex]. Therefore, we can describe the sequence by the explicit formula [latex]{a}_{n}=\\frac{{3}^{n}}{3n+1}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739016124\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169739223015\" data-type=\"exercise\">\n<div id=\"fs-id1169736593628\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169736593628\" data-type=\"problem\">\n<p id=\"fs-id1169739242722\">Find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence [latex]\\left\\{\\frac{1}{5},-\\frac{1}{7},\\frac{1}{9},-\\frac{1}{11}\\text{,}\\ldots\\right\\}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738880222\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739045640\">The denominators form an arithmetic sequence.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739093464\" data-type=\"solution\">\n<p id=\"fs-id1169738966727\">[latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n+1}}{3+2n}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oduNoQPFDD4?controls=0&amp;start=285&amp;end=348&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.1.1_285to348_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.1&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1169739010866\" data-type=\"example\">\n<div id=\"fs-id1169738890687\" data-type=\"exercise\">\n<div id=\"fs-id1169739182834\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Defined by Recurrence Relations<\/h3>\n<div id=\"fs-id1169739182834\" data-type=\"problem\">\n<p id=\"fs-id1169736617641\">For each of the following recursively defined sequences, find an explicit formula for the sequence.<\/p>\n<ol id=\"fs-id1169739350728\" type=\"a\">\n<li>[latex]{a}_{1}=2[\/latex], [latex]{a}_{n}=-3{a}_{n - 1}[\/latex] for [latex]n\\ge 2[\/latex]<\/li>\n<li>[latex]{a}_{1}=\\frac{1}{2}[\/latex], [latex]{a}_{n}={a}_{n - 1}+{\\left(\\frac{1}{2}\\right)}^{n}[\/latex] for [latex]n\\ge 2[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739027829\" data-type=\"solution\">\n<ol id=\"fs-id1169738999234\" type=\"a\">\n<li>Writing out the first few terms, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738860999\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{a}_{1}=2\\hfill \\\\ {a}_{2}=-3{a}_{1}=-3\\left(2\\right)\\hfill \\\\ {a}_{3}=-3{a}_{2}={\\left(-3\\right)}^{2}2\\hfill \\\\ {a}_{4}=-3{a}_{3}={\\left(-3\\right)}^{3}2.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIn general,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736709052\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}=2{\\left(-3\\right)}^{n - 1}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Write out the first few terms:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738999519\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ {a}_{1}=\\frac{1}{2}\\hfill \\\\ {a}_{2}={a}_{1}+{\\left(\\frac{1}{2}\\right)}^{2}=\\frac{1}{2}+\\frac{1}{4}=\\frac{3}{4}\\hfill \\\\ {a}_{3}={a}_{2}+{\\left(\\frac{1}{2}\\right)}^{3}=\\frac{3}{4}+\\frac{1}{8}=\\frac{7}{8}\\hfill \\\\ {a}_{4}={a}_{3}+{\\left(\\frac{1}{2}\\right)}^{4}=\\frac{7}{8}+\\frac{1}{16}=\\frac{15}{16}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFrom this pattern, we derive the explicit formula<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739019569\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}=\\frac{{2}^{n}-1}{{2}^{n}}=1-\\frac{1}{{2}^{n}}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739016786\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169738936238\" data-type=\"exercise\">\n<div id=\"fs-id1169738975437\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169738975437\" data-type=\"problem\">\n<p id=\"fs-id1169739214209\">Find an explicit formula for the sequence defined recursively such that [latex]{a}_{1}=-4[\/latex] and [latex]{a}_{n}={a}_{n - 1}+6[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739302833\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739102236\">This is an arithmetic sequence.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736854641\" data-type=\"solution\">\n<p id=\"fs-id1169739242787\">[latex]{a}_{n}=6n - 10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oduNoQPFDD4?controls=0&amp;start=475&amp;end=506&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.1.1_475to506_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.1&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm19431\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19431&theme=oea&iframe_resize_id=ohm19431&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1169738981819\" data-depth=\"1\"><\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1692\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>5.1.1. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.1.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1692","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1692","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1692\/revisions"}],"predecessor-version":[{"id":2210,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1692\/revisions\/2210"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1692\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1692"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1692"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1692"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1692"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}