{"id":859,"date":"2025-06-20T17:18:31","date_gmt":"2025-06-20T17:18:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=859"},"modified":"2025-09-11T15:49:11","modified_gmt":"2025-09-11T15:49:11","slug":"basics-of-differential-equations-learn-it-1-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations-learn-it-1-2\/","title":{"raw":"Sequences and Their Properties: Learn It 1","rendered":"Sequences and Their Properties: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find the pattern in a sequence and write a formula for its terms<\/li>\r\n \t<li>Determine what value a sequence approaches (if it approaches any value at all)<\/li>\r\n \t<li>Figure out if a sequence converges or diverges<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>What is a Sequence?<\/h2>\r\nA sequence is simply an ordered list of numbers that follows a pattern. You encounter sequences regularly\u2014think of your monthly rent payments, the distances you run each week, or even the Fibonacci numbers. Understanding sequences helps us analyze patterns and predict future behavior in mathematical models.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>infinite sequence<\/h3>\r\n<p class=\"whitespace-normal break-words\">An <strong>infinite sequence<\/strong> [latex]{a_n}[\/latex] is an ordered list of numbers:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]a_1, a_2, a_3, \\ldots, a_n, \\ldots[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Key terms:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Each number [latex]a_n[\/latex] is called a <strong>term<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">The subscript [latex]n[\/latex] is the <strong>index variable<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">We write sequences as [latex]{a_n}_{n=1}^{\\infty}[\/latex] or simply [latex]{a_n}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Important:<\/strong> A similar notation is used for sets, but a sequence is an ordered list (like a line at the grocery store), while a set is unordered collection (like items in a shopping basket). Order matters for sequences!<\/section>\r\n<p class=\"whitespace-normal break-words\">Consider this sequence: [latex]2, 4, 8, 16, 32, \\ldots[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">You can see that [latex]a_1 = 2[\/latex], [latex]a_2 = 4[\/latex], [latex]a_3 = 8[\/latex], and so on. Each term is a power of 2. But how do we capture this pattern mathematically? There are two main approaches.<\/p>\r\n<p class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\"><strong>Method 1: Explicit Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">An <strong>explicit formula<\/strong> gives you [latex]a_n[\/latex] directly in terms of [latex]n[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">For our sequence: [latex]a_n = 2^n[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This means [latex]a_1 = 2^1 = 2[\/latex], [latex]a_2 = 2^2 = 4[\/latex], [latex]a_3 = 2^3 = 8[\/latex], and so on. The beauty of an explicit formula is that you can find any term directly. Want the 100th term? Just calculate [latex]a_{100} = 2^{100}[\/latex] without computing all the previous terms.<\/p>\r\n<p class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\"><strong>Method 2: Recurrence Relation<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A <strong>recurrence relation<\/strong> defines each term using previous terms.<\/p>\r\n<p class=\"whitespace-normal break-words\">For the same sequence: [latex]a_1 = 2[\/latex] and [latex]a_n = 2a_{n-1}[\/latex] for [latex]n \\geq 2[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This says \"start with 2, then each new term is twice the previous term.\"<\/p>\r\nRecurrence relations are particularly useful when each term naturally depends on what came before\u2014like population growth where next year's population depends on this year's. However, finding a specific term requires calculating all the previous ones.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Sequences don't have to start at [latex]n = 1[\/latex]. You might see sequences starting at [latex]n = 0[\/latex] or any other integer, depending on the context. Just pay attention to the index range given.<\/section>\r\n<p class=\"whitespace-normal break-words\">Since a sequence [latex]{a_n}[\/latex] assigns exactly one value to each positive integer [latex]n[\/latex], we can think of it as a function whose domain is the set of positive integers. This means we can graph sequences just like we graph functions.<\/p>\r\n<p class=\"whitespace-normal break-words\">The graph of sequence [latex]{a_n}[\/latex] consists of discrete points [latex](n, a_n)[\/latex] for all positive integers [latex]n[\/latex]. Unlike continuous functions, sequence graphs show isolated dots rather than connected curves.<\/p>\r\n\r\n[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]\r\n<h2>Special Types of Sequences<\/h2>\r\nTwo types of sequences appear frequently in mathematics and have special names because of their predictable patterns.\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Arithmetic Sequences<\/h3>\r\nIn an\u00a0<strong>arithmetic sequence<\/strong>, the\u00a0<em data-effect=\"italics\">difference<\/em>\u00a0between every pair of consecutive terms is the same.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3><strong>arithmetic sequence<\/strong><\/h3>\r\n<p class=\"whitespace-normal break-words\">An <strong>arithmetic sequence<\/strong> has the same difference between consecutive terms.<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">General form: [latex]a_n = cn + b[\/latex] (explicit formula)<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">The common difference is [latex]d = a_{n+1} - a_n[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">Consider the sequence [latex]3, 7, 11, 15, 19, \\ldots[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Notice that each term increases by 4: [latex]7-3=4[\/latex], [latex]11-7=4[\/latex], [latex]15-11=4[\/latex], and so on.<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">We can describe this sequence using a recurrence relation: [latex]a_1 = 3[\/latex] and [latex]a_n = a_{n-1} + 4[\/latex] for [latex]n \\geq 2[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Or we can find the explicit formula by observing the pattern:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_2 = 3 + 4[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_3 = 3 + 4 + 4 = 3 + 2 \\cdot 4[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_4 = 3 + 4 + 4 + 4 = 3 + 3 \\cdot 4[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">This gives us the explicit formula: [latex]a_n = 3 + 4(n-1) = 4n - 1[\/latex]<\/p>\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Geometric Sequences<\/h3>\r\nIn a\u00a0<strong>geometric sequence<\/strong>, the\u00a0<em data-effect=\"italics\">ratio<\/em>\u00a0of every pair of consecutive terms is the same.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3><strong>geometric sequence<\/strong><\/h3>\r\n<p class=\"whitespace-normal break-words\">A <strong>geometric sequence<\/strong> has the same ratio between consecutive terms.<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">General form: [latex]a_n = cr^n[\/latex] (explicit formula)<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">The common ratio is [latex]r = \\frac{a_{n+1}}{a_n}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">Consider the sequence [latex]2, -\\frac{2}{3}, \\frac{2}{9}, -\\frac{2}{27}, \\frac{2}{81}, \\ldots[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Each term is [latex]-\\frac{1}{3}[\/latex] times the previous term. We can verify: [latex]\\frac{-\\frac{2}{3}}{2} = -\\frac{1}{3}[\/latex] and [latex]\\frac{\\frac{2}{9}}{-\\frac{2}{3}} = -\\frac{1}{3}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Using a recurrence relation: [latex]a_1 = 2[\/latex] and [latex]a_n = -\\frac{1}{3} \\cdot a_{n-1}[\/latex] for [latex]n \\geq 2[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Or we can find the explicit formula by following the pattern:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_2 = -\\frac{1}{3} \\cdot 2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_3 = \\left(-\\frac{1}{3}\\right)^2 \\cdot 2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_4 = \\left(-\\frac{1}{3}\\right)^3 \\cdot 2[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">This gives us: [latex]a_n = 2\\left(-\\frac{1}{3}\\right)^{n-1}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\"><strong>Quick Recognition<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Arithmetic:<\/strong> Look for the same number being added (or subtracted) repeatedly<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Geometric:<\/strong> Look for the same number being multiplied repeatedly<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311393[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find the pattern in a sequence and write a formula for its terms<\/li>\n<li>Determine what value a sequence approaches (if it approaches any value at all)<\/li>\n<li>Figure out if a sequence converges or diverges<\/li>\n<\/ul>\n<\/section>\n<h2>What is a Sequence?<\/h2>\n<p>A sequence is simply an ordered list of numbers that follows a pattern. You encounter sequences regularly\u2014think of your monthly rent payments, the distances you run each week, or even the Fibonacci numbers. Understanding sequences helps us analyze patterns and predict future behavior in mathematical models.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>infinite sequence<\/h3>\n<p class=\"whitespace-normal break-words\">An <strong>infinite sequence<\/strong> [latex]{a_n}[\/latex] is an ordered list of numbers:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]a_1, a_2, a_3, \\ldots, a_n, \\ldots[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Key terms:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Each number [latex]a_n[\/latex] is called a <strong>term<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">The subscript [latex]n[\/latex] is the <strong>index variable<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">We write sequences as [latex]{a_n}_{n=1}^{\\infty}[\/latex] or simply [latex]{a_n}[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Important:<\/strong> A similar notation is used for sets, but a sequence is an ordered list (like a line at the grocery store), while a set is unordered collection (like items in a shopping basket). Order matters for sequences!<\/section>\n<p class=\"whitespace-normal break-words\">Consider this sequence: [latex]2, 4, 8, 16, 32, \\ldots[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">You can see that [latex]a_1 = 2[\/latex], [latex]a_2 = 4[\/latex], [latex]a_3 = 8[\/latex], and so on. Each term is a power of 2. But how do we capture this pattern mathematically? There are two main approaches.<\/p>\n<p class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\"><strong>Method 1: Explicit Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">An <strong>explicit formula<\/strong> gives you [latex]a_n[\/latex] directly in terms of [latex]n[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">For our sequence: [latex]a_n = 2^n[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This means [latex]a_1 = 2^1 = 2[\/latex], [latex]a_2 = 2^2 = 4[\/latex], [latex]a_3 = 2^3 = 8[\/latex], and so on. The beauty of an explicit formula is that you can find any term directly. Want the 100th term? Just calculate [latex]a_{100} = 2^{100}[\/latex] without computing all the previous terms.<\/p>\n<p class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\"><strong>Method 2: Recurrence Relation<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A <strong>recurrence relation<\/strong> defines each term using previous terms.<\/p>\n<p class=\"whitespace-normal break-words\">For the same sequence: [latex]a_1 = 2[\/latex] and [latex]a_n = 2a_{n-1}[\/latex] for [latex]n \\geq 2[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This says &#8220;start with 2, then each new term is twice the previous term.&#8221;<\/p>\n<p>Recurrence relations are particularly useful when each term naturally depends on what came before\u2014like population growth where next year&#8217;s population depends on this year&#8217;s. However, finding a specific term requires calculating all the previous ones.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Sequences don&#8217;t have to start at [latex]n = 1[\/latex]. You might see sequences starting at [latex]n = 0[\/latex] or any other integer, depending on the context. Just pay attention to the index range given.<\/section>\n<p class=\"whitespace-normal break-words\">Since a sequence [latex]{a_n}[\/latex] assigns exactly one value to each positive integer [latex]n[\/latex], we can think of it as a function whose domain is the set of positive integers. This means we can graph sequences just like we graph functions.<\/p>\n<p class=\"whitespace-normal break-words\">The graph of sequence [latex]{a_n}[\/latex] consists of discrete points [latex](n, a_n)[\/latex] for all positive integers [latex]n[\/latex]. Unlike continuous functions, sequence graphs show isolated dots rather than connected curves.<\/p>\n<figure style=\"width: 487px\" 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\" \/><figcaption class=\"wp-caption-text\">Figure 1. The plotted points are a graph of the sequence [latex]\\left\\{{2}^{n}\\right\\}[\/latex].<\/figcaption><\/figure>\n<h2>Special Types of Sequences<\/h2>\n<p>Two types of sequences appear frequently in mathematics and have special names because of their predictable patterns.<\/p>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Arithmetic Sequences<\/h3>\n<p>In an\u00a0<strong>arithmetic sequence<\/strong>, the\u00a0<em data-effect=\"italics\">difference<\/em>\u00a0between every pair of consecutive terms is the same.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3><strong>arithmetic sequence<\/strong><\/h3>\n<p class=\"whitespace-normal break-words\">An <strong>arithmetic sequence<\/strong> has the same difference between consecutive terms.<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">General form: [latex]a_n = cn + b[\/latex] (explicit formula)<\/p>\n<p class=\"whitespace-pre-wrap break-words\">The common difference is [latex]d = a_{n+1} - a_n[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">Consider the sequence [latex]3, 7, 11, 15, 19, \\ldots[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Notice that each term increases by 4: [latex]7-3=4[\/latex], [latex]11-7=4[\/latex], [latex]15-11=4[\/latex], and so on.<\/p>\n<p class=\"whitespace-pre-wrap break-words\">We can describe this sequence using a recurrence relation: [latex]a_1 = 3[\/latex] and [latex]a_n = a_{n-1} + 4[\/latex] for [latex]n \\geq 2[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Or we can find the explicit formula by observing the pattern:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]a_2 = 3 + 4[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]a_3 = 3 + 4 + 4 = 3 + 2 \\cdot 4[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]a_4 = 3 + 4 + 4 + 4 = 3 + 3 \\cdot 4[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">This gives us the explicit formula: [latex]a_n = 3 + 4(n-1) = 4n - 1[\/latex]<\/p>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Geometric Sequences<\/h3>\n<p>In a\u00a0<strong>geometric sequence<\/strong>, the\u00a0<em data-effect=\"italics\">ratio<\/em>\u00a0of every pair of consecutive terms is the same.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3><strong>geometric sequence<\/strong><\/h3>\n<p class=\"whitespace-normal break-words\">A <strong>geometric sequence<\/strong> has the same ratio between consecutive terms.<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">General form: [latex]a_n = cr^n[\/latex] (explicit formula)<\/p>\n<p class=\"whitespace-pre-wrap break-words\">The common ratio is [latex]r = \\frac{a_{n+1}}{a_n}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">Consider the sequence [latex]2, -\\frac{2}{3}, \\frac{2}{9}, -\\frac{2}{27}, \\frac{2}{81}, \\ldots[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Each term is [latex]-\\frac{1}{3}[\/latex] times the previous term. We can verify: [latex]\\frac{-\\frac{2}{3}}{2} = -\\frac{1}{3}[\/latex] and [latex]\\frac{\\frac{2}{9}}{-\\frac{2}{3}} = -\\frac{1}{3}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Using a recurrence relation: [latex]a_1 = 2[\/latex] and [latex]a_n = -\\frac{1}{3} \\cdot a_{n-1}[\/latex] for [latex]n \\geq 2[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Or we can find the explicit formula by following the pattern:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]a_2 = -\\frac{1}{3} \\cdot 2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]a_3 = \\left(-\\frac{1}{3}\\right)^2 \\cdot 2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]a_4 = \\left(-\\frac{1}{3}\\right)^3 \\cdot 2[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">This gives us: [latex]a_n = 2\\left(-\\frac{1}{3}\\right)^{n-1}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\"><strong>Quick Recognition<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Arithmetic:<\/strong> Look for the same number being added (or subtracted) repeatedly<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Geometric:<\/strong> Look for the same number being multiplied repeatedly<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558896\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311393\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311393&theme=lumen&iframe_resize_id=ohm311393&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":671,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/859"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/859\/revisions"}],"predecessor-version":[{"id":2315,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/859\/revisions\/2315"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/859\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=859"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=859"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=859"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=859"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}