{"id":1856,"date":"2023-10-12T00:32:20","date_gmt":"2023-10-12T00:32:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-sequences-and-their-notations\/"},"modified":"2023-10-12T00:32:20","modified_gmt":"2023-10-12T00:32:20","slug":"introduction-sequences-and-their-notations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-sequences-and-their-notations\/","title":{"raw":"Sequences and Their Notations","rendered":"Sequences and Their Notations"},"content":{"raw":"\n\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul class=\"ul1\">\n \t<li class=\"li2\"><span class=\"s1\">Write the terms of a sequence defined by an explicit formula.<\/span><\/li>\n \t<li class=\"li2\"><span class=\"s1\">Write the terms of a sequence defined by a recursive formula.<\/span><\/li>\n \t<li class=\"li2\"><span class=\"s1\">Use factorial notation.<\/span><\/li>\n<\/ul>\n<\/div>\nA video game company launches an exciting new advertising campaign. They predict the number of online visits to their website, or hits, will double each day. The model they are using shows 2 hits the first day, 4 hits the second day, 8 hits the third day, and so on.\n<table>\n<tbody>\n<tr>\n<td><strong>Day<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>\u2026<\/td>\n<\/tr>\n<tr>\n<td><strong>Hits<\/strong><\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<td>16<\/td>\n<td>32<\/td>\n<td>\u2026<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nIf their model continues, how many hits will there be at the end of the month? To answer this question, we\u2019ll first need to know how to determine a list of numbers written in a specific order. In this section we will explore these kinds of ordered lists.\n<h2>Sequences Defined by an Explicit Formula<\/h2>\nOne way to describe an ordered list of numbers is as a <strong>sequence<\/strong>. A sequence is a function whose domain is a subset of the counting numbers. The sequence established by the number of hits on the website is\n<p style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots \\right\\}[\/latex].<\/p>\nThe <strong>ellipsis<\/strong> (\u2026) indicates that the sequence continues indefinitely. Each number in the sequence is called a <strong>term<\/strong>. The first five terms of this sequence are 2, 4, 8, 16, and 32.\n\nListing all of the terms for a sequence can be cumbersome. For example, finding the number of hits on the website at the end of the month would require listing out as many as 31 terms. A more efficient way to determine a specific term is by writing a formula to define the sequence.\n\nOne type of formula is an <strong>explicit formula<\/strong>, which defines the terms of a sequence using their position in the sequence. Explicit formulas are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the <strong> [latex]n\\text{th}[\/latex] term of the sequence<\/strong>, where [latex]n[\/latex] is any positive number. In our example, each number in the sequence is double the previous number, so we can use powers of 2 to write a formula for the [latex]n\\text{th}[\/latex] term.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03220152\/CNX_Precalc_Figure_11_01_0012.jpg\" alt=\"Sequence of {2, 4, 8, 16, 32, ...} expressed in exponential form (i.e., {2^1, 2^2, 2^3, ..., 2^n, ...}\" width=\"487\" height=\"89\">\n\nThe first term of the sequence is [latex]{2}^{1}=2[\/latex], the second term is [latex]{2}^{2}=4[\/latex], the third term is [latex]{2}^{3}=8[\/latex], and so on. The [latex]n\\text{th}[\/latex] term of the sequence can be found by raising 2 to the [latex]n\\text{th}[\/latex] power. An explicit formula for a sequence is named by a lower case letter [latex]a,b,c..[\/latex]. with the subscript [latex]n[\/latex]. The explicit formula for this sequence is\n<p style=\"text-align: center;\">[latex]{a}_{n}={2}^{n}[\/latex]<\/p>\nNow that we have a formula for the [latex]n\\text{th}[\/latex] term of the sequence, we can answer the question posed at the beginning of this section. We were asked to find the number of hits at the end of the month, which we will take to be 31 days. To find the number of hits on the last day of the month, we need to find the 31<sup>st<\/sup> term of the sequence. We will substitute 31 for [latex]n[\/latex] in the formula.\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{31}&amp;={2}^{31} \\\\ &amp;=\\text{2,147,483,648} \\end{align}[\/latex]<\/p>\nIf the doubling trend continues, the company will get [latex]\\text{2,147,483,648}[\/latex] hits on the last day of the month. That is over 2.1 billion hits! The huge number is probably a little unrealistic because it does not take consumer interest and competition into account. It does, however, give the company a starting point from which to consider business decisions.\n\nAnother way to represent the sequence is by using a table. The first five terms of the sequence and the [latex]n\\text{th}[\/latex] term of the sequence are shown in the table.\n<table>\n<tbody>\n<tr>\n<td><strong> [latex]n[\/latex] <\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>[latex]n[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]n\\text{th}[\/latex] term of the sequence, [latex]{a}_{n}[\/latex] <\/strong><\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<td>16<\/td>\n<td>32<\/td>\n<td>[latex]{2}^{n}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nGraphing provides a visual representation of the sequence as a set of distinct points. We can see from the graph below that the number of hits is rising at an exponential rate. This particular sequence forms an exponential function.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03220155\/CNX_Precalc_Figure_11_01_0022.jpg\" alt=\"Graph of a plotted exponential function, f(n) = 2^n, where the x-axis is labeled n and the y-axis is labeled a_n.\" width=\"487\" height=\"397\">\n\nLastly, we can write this particular sequence as\n<p style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots ,{2}^{n},\\dots \\right\\}[\/latex].<\/p>\nA sequence that continues indefinitely is called an <strong>infinite sequence<\/strong>. The domain of an infinite sequence is the set of counting numbers. If we consider only the first 10 terms of the sequence, we could write\n<p style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots ,1024\\right\\}[\/latex].<\/p>\nThis sequence is called a <strong>finite sequence<\/strong> because it does not continue indefinitely.\n<div class=\"textbox\">\n<h3>A General Note: Sequence<\/h3>\nA <strong>sequence<\/strong> is a function whose domain is the set of positive integers. A <strong>finite sequence<\/strong> is a sequence whose domain consists of only the first [latex]n[\/latex] positive integers. The numbers in a sequence are called <strong>terms<\/strong>. The variable [latex]a[\/latex] with a number subscript is used to represent the terms in a sequence and to indicate the position of the term in the sequence.\n<p style=\"text-align: center;\">[latex]{a}_{1},{a}_{2},{a}_{3},\\dots ,{a}_{n},\\dots [\/latex]<\/p>\nWe call [latex]{a}_{1}[\/latex] the first term of the sequence, [latex]{a}_{2}[\/latex] the second term of the sequence, [latex]{a}_{3}[\/latex] the third term of the sequence, and so on. The term [latex]{a}_{n}[\/latex] is called the <strong> [latex]n\\text{th}[\/latex] term of the sequence<\/strong>, or the general term of the sequence. An <strong>explicit formula<\/strong> defines the [latex]n\\text{th}[\/latex] term of a sequence using the position of the term. A sequence that continues indefinitely is an <strong>infinite sequence<\/strong>.\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Does a sequence always have to begin with [latex]{a}_{1}?[\/latex]<\/h3>\n<em>No. In certain problems, it may be useful to define the initial term as [latex]{a}_{0}[\/latex] instead of [latex]{a}_{1}[\/latex]. In these problems, the domain of the function includes 0.<\/em>\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an explicit formula, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\n<ol>\n \t<li>Substitute each value of [latex]n[\/latex] into the formula. Begin with [latex]n=1[\/latex] to find the first term, [latex]{a}_{1}[\/latex].<\/li>\n \t<li>To find the second term, [latex]{a}_{2}[\/latex], use [latex]n=2[\/latex].<\/li>\n \t<li>Continue in the same manner until you have identified all [latex]n[\/latex] terms.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Terms of a Sequence Defined by an Explicit Formula<\/h3>\nWrite the first five terms of the sequence defined by the explicit formula [latex]{a}_{n}=-3n+8[\/latex].\n\n[reveal-answer q=\"776421\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"776421\"]\n\nSubstitute [latex]n=1[\/latex] into the formula. Repeat with values 2 through 5 for [latex]n[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{lign}n=1 &amp;&amp; {a}_{1}=-3\\left(1\\right)+8=5 \\\\ n=2 &amp;&amp; {a}_{2}=-3\\left(2\\right)+8=2 \\\\ n=3 &amp;&amp; {a}_{3}=-3\\left(3\\right)+8=-1 \\\\ n=4 &amp;&amp; {a}_{4}=-3\\left(4\\right)+8=-4 \\\\ n=5 &amp;&amp; {a}_{5}=-3\\left(5\\right)+8=-7 \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">The first five terms are [latex]\\left\\{5,2,-1,-4,-7\\right\\}[\/latex].<\/p>\n\n<h4>Analysis of the Solution<\/h4>\nThe sequence values can be listed in a table. A table&nbsp;is a convenient way to input the function into a graphing utility.\n<table summary=\"\">\n<tbody>\n<tr>\n<td><strong> [latex]n[\/latex] <\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]{a}_{n}[\/latex] <\/strong><\/td>\n<td>5<\/td>\n<td>2<\/td>\n<td>\u20131<\/td>\n<td>\u20134<\/td>\n<td>\u20137<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nA graph can be made from this table of values. From the graph in Figure 2, we can see that this sequence represents a linear function, but notice the graph is not continuous because the domain is over the positive integers only.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03220157\/CNX_Precalc_Figure_11_01_0032.jpg\" alt=\"Graph of a scattered plot where the x-axis is labeled n and the y-axis is labeled a_n.\" width=\"487\" height=\"584\">\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nWrite the first five terms of the sequence defined by the <strong>explicit formula<\/strong> [latex]{t}_{n}=5n - 4[\/latex].\n\n[reveal-answer q=\"25802\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"25802\"]\n\nThe first five terms are [latex]\\left\\{1,6, 11, 16, 21\\right\\}[\/latex]\n\n[\/hidden-answer]\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5846&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe>\n\n<\/div>\n<h2>Finding an Explicit Formula<\/h2>\nThus far, we have been given the explicit formula and asked to find a number of terms of the sequence. Sometimes, the explicit formula for the [latex]n\\text{th}[\/latex] term of a sequence is not given. Instead, we are given several terms from the sequence. When this happens, we can work in reverse to find an explicit formula from the first few terms of a sequence. The key to finding an explicit formula is to look for a pattern in the terms. Keep in mind that the pattern may involve alternating terms, formulas for numerators, formulas for denominators, exponents, or bases.\n<div class=\"textbox\">\n<h3>How To: Given the first few terms of a sequence, find an explicit formula for the sequence.<\/h3>\n<ol>\n \t<li>Look for a pattern among the terms.<\/li>\n \t<li>If the terms are fractions, look for a separate pattern among the numerators and denominators.<\/li>\n \t<li>Look for a pattern among the signs of the terms.<\/li>\n \t<li>Write a formula for [latex]{a}_{n}[\/latex] in terms of [latex]n[\/latex]. Test your formula for [latex]n=1,n=2[\/latex],&nbsp;and&nbsp;[latex]n=3[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing an Explicit Formula for the <em style=\"font-size: 16px;\">n<\/em><span style=\"font-size: 16px;\">th Term of a Sequence<\/span><\/h3>\nWrite an explicit formula for the [latex]n\\text{th}[\/latex] term of each sequence.\n<ol>\n \t<li>[latex]\\left\\{-\\dfrac{2}{11},\\dfrac{3}{13},-\\dfrac{4}{15},\\dfrac{5}{17},-\\dfrac{6}{19},\\dots \\right\\}[\/latex]<\/li>\n \t<li>[latex]\\left\\{-\\dfrac{2}{25},-\\dfrac{2}{125},-\\dfrac{2}{625},-\\dfrac{2}{3\\text{,}125},-\\dfrac{2}{15\\text{,}625},\\dots \\right\\}[\/latex]<\/li>\n \t<li>[latex]\\left\\{{e}^{4},{e}^{5},{e}^{6},{e}^{7},{e}^{8},\\dots \\right\\}[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"328784\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"328784\"]\n\nLook for the pattern in each sequence.\n<ol>\n \t<li>The terms alternate between positive and negative. We can use [latex]{\\left(-1\\right)}^{n}[\/latex] to make the terms alternate. The numerator can be represented by [latex]n+1[\/latex]. The denominator can be represented by [latex]2n+9[\/latex].\n<div style=\"text-align: center;\">[latex]{a}_{n}=\\dfrac{{\\left(-1\\right)}^{n}\\left(n+1\\right)}{2n+9}[\/latex]<\/div><\/li>\n \t<li>The terms are all negative.\n<div style=\"text-align: center;\">[latex]\\begin{align} &amp; \\left\\{-\\dfrac{2}{25},-\\dfrac{2}{125},-\\dfrac{2}{625},-\\dfrac{2}{3\\text{,}125},-\\dfrac{2}{15\\text{,}625},\\dots \\right\\}&amp;&amp; \\text{The numerator is 2.} \\\\[1mm] &amp; \\left\\{-\\dfrac{2}{5^2},-\\dfrac{2}{5^3},-\n\\dfrac{2}{5^4},-\\dfrac{2}{5^5},-\\dfrac{2}{5^6},\\dots,-\\dfrac{2}{5^n},\\dots \\right\\} &amp;&amp; \\text{The denominators are increasing powers of 5.}\\end{align}[\/latex]<\/div>\nSo we know that the fraction is negative, the numerator is 2, and the denominator can be represented by [latex]{5}^{n+1}[\/latex].\n<div style=\"text-align: center;\">[latex]{a}_{n}=-\\dfrac{2}{{5}^{n+1}}[\/latex]<\/div><\/li>\n \t<li>The terms are powers of [latex]e[\/latex]. For [latex]n=1[\/latex], the first term is [latex]{e}^{4}[\/latex] so the exponent must be [latex]n+3[\/latex].\n<div style=\"text-align: center;\">[latex]{a}_{n}={e}^{n+3}[\/latex]<\/div><\/li>\n<\/ol>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nWrite an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence.\n<p style=\"text-align: center;\">[latex]\\{9;\u221281,729;\u22126,561;59,049\\}[\/latex]<\/p>\n[reveal-answer q=\"806943\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"806943\"]\n\n[latex]{a}_{n}={\\left(-1\\right)}^{n+1}{9}^{n}[\/latex]\n\n[\/hidden-answer]\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5805&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"550\"><\/iframe>\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nWrite an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence.\n<div style=\"text-align: center;\">[latex]\\left\\{-\\dfrac{3}{4},-\\dfrac{9}{8},-\\dfrac{27}{12},-\\dfrac{81}{16},-\\dfrac{243}{20},\\dots\\right\\}[\/latex]<\/div>\n[reveal-answer q=\"282988\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"282988\"]\n\n[latex]{a}_{n}=-\\dfrac{{3}^{n}}{4n}[\/latex]\n\n[\/hidden-answer]\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5806&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"550\"><\/iframe>\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nWrite an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence.\n<div style=\"text-align: center;\">[latex]\\left\\{\\dfrac{1}{{e}^{2}}, \\dfrac{1}{e}, 1, e, {e}^{2},...\\right\\}[\/latex]<\/div>\n[reveal-answer q=\"804020\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"804020\"]\n\n[latex]{a}_{n}={e}^{n - 3}[\/latex]\n\n[\/hidden-answer]\n\n<\/div>\n<h2>Investigating Alternating Sequences<\/h2>\nSometimes sequences have terms that alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. Let\u2019s take a look at the following sequence.\n<div style=\"text-align: center;\">[latex]\\left\\{2,-4,6,-8\\right\\}[\/latex]<\/div>\nNotice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence.\n<div class=\"textbox\">\n<h3>How To: Given an explicit formula with alternating terms, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\n<ol>\n \t<li>Substitute each value of [latex]n[\/latex] into the formula. Begin with [latex]n=1[\/latex] to find the first term, [latex]{a}_{1}[\/latex]. The sign of the term in the explicit formula is given by the [latex]{\\left(-1\\right)}^{n}[\/latex] if the first term is negative and&nbsp;[latex]{\\left(-1\\right)}^{n-1}[\/latex] if the first term is positive.<\/li>\n \t<li>To find the second term, [latex]{a}_{2}[\/latex], use [latex]n=2[\/latex].<\/li>\n \t<li>Continue in the same manner until you have identified all [latex]n[\/latex] terms.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Terms of an Alternating Sequence Defined by an Explicit Formula<\/h3>\nWrite the first five terms of the sequence.\n<div style=\"text-align: center;\">[latex]{a}_{n}=\\dfrac{{\\left(-1\\right)}^{n}{n}^{2}}{n+1}[\/latex]<\/div>\n[reveal-answer q=\"979322\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"979322\"]\n\nSubstitute [latex]n=1[\/latex], [latex]n=2[\/latex], and so on in the formula.\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;n=1 &amp;&amp; {a}_{1}=\\dfrac{{\\left(-1\\right)}^{1}{2}^{2}}{1+1}=-\\dfrac{1}{2} \\\\[1mm] &amp;n=2 &amp;&amp; {a}_{2}=\\dfrac{{\\left(-1\\right)}^{2}{2}^{2}}{2+1}=\\dfrac{4}{3} \\\\[1mm] &amp;n=3 &amp;&amp; {a}_{3}=\\dfrac{{\\left(-1\\right)}^{3}{3}^{2}}{3+1}=-\\frac{9}{4} \\\\[1mm] &amp;n=4 &amp;&amp; {a}_{4}=\\dfrac{{\\left(-1\\right)}^{4}{4}^{2}}{4+1}=\\dfrac{16}{5} \\\\[1mm] &amp;n=5 &amp;&amp; {a}_{5}=\\dfrac{{\\left(-1\\right)}^{5}{5}^{2}}{5+1}=-\\dfrac{25}{6} \\end{align}[\/latex]<\/div>\n<div><\/div>\nThe first five terms are [latex]\\left\\{-\\dfrac{1}{2},\\dfrac{4}{3},-\\dfrac{9}{4},\\dfrac{16}{5},-\\dfrac{25}{6}\\right\\}[\/latex].\n<h4>Analysis of the Solution<\/h4>\nThe graph of this function looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03220159\/CNX_Precalc_Figure_11_01_0042.jpg\" alt=\"Graph of a scattered plot with labeled points: (1, -1\/2), (2, 4\/3), (3, -9\/4), (4, 16\/5), and (5, -25\/6). The x-axis is labeled n and the y-axis is labeled a_n.\" width=\"487\" height=\"482\">\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>In Example: Writing the Terms of an Alternating Sequence Defined by an Explicit Formula, does the (\u20131) to the power of [latex]n[\/latex] account for the oscillations of signs?<\/h4>\n<em>Yes, the power might be<\/em> [latex]n,n+1,n - 1[\/latex], <em>and so on, but any odd powers will result in a negative term, and any even power will result in a positive term.<\/em>\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nWrite the first five terms of the sequence:\n<div style=\"text-align: center;\">[latex]{a}_{n}=\\dfrac{4n}{{\\left(-2\\right)}^{n}}[\/latex]<\/div>\n<div><\/div>\n[reveal-answer q=\"196840\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"196840\"]\n\nThe first five terms are [latex]\\left\\{-2, 2, -\\dfrac{3}{2}, 1,-\\dfrac{5}{8}\\right\\}[\/latex].\n\n[\/hidden-answer]\n<iframe id=\"mom30\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5823&amp;theme=oea&amp;iframe_resize_id=mom30\" width=\"100%\" height=\"350\"><\/iframe>\n\n<\/div>\n<h2>Sequences Defined by a Recursive Formula<\/h2>\nSequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34,\u2026. Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with 13 petals, and different varieties of daisies that may have 21 or 34 petals.\n\nEach term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a <strong>recursive formula<\/strong>, a formula that defines the terms of a sequence using previous terms.\n\nA recursive formula always has two parts: the value of an initial term (or terms), and an equation defining [latex]{a}_{n}[\/latex] in terms of preceding terms. For example, suppose we know the following:\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&amp;=3 \\\\ {a}_{n}&amp;=2{a}_{n - 1}-1, \\text{for } n\\ge 2 \\end{align}[\/latex]<\/p>\nWe can find the subsequent terms of the sequence using the first term.\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&amp;=3\\\\ {a}_{2}&amp;=2{a}_{1}-1=2\\left(3\\right)-1=5\\\\ {a}_{3}&amp;=2{a}_{2}-1=2\\left(5\\right)-1=9\\\\ {a}_{4}&amp;=2{a}_{3}-1=2\\left(9\\right)-1=17\\end{align}[\/latex]<\/p>\nSo the first four terms of the sequence are [latex]\\left\\{3,5,9,17\\right\\}[\/latex].\n\nThe recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms.\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&amp;=1 \\\\ {a}_{2}&amp;=1 \\\\ {a}_{n}&amp;={a}_{n - 1}+{a}_{n - 2}, \\text{for } n\\ge 3 \\end{align}[\/latex]<\/p>\nTo find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We saw above that the eighth and ninth terms are 21 and 34, so\n<p style=\"text-align: center;\">[latex]{a}_{10}={a}_{9}+{a}_{8}=34+21=55[\/latex]<\/p>\n\n<div class=\"textbox\">\n<h3>A General Note: Recursive Formula<\/h3>\nA <strong>recursive formula<\/strong> is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence.\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Must the first two terms always be given in a recursive formula?<\/h4>\n<em>No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only the first term to be defined.<\/em>\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a recursive formula with only the first term provided, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\n<ol>\n \t<li>Identify the initial term, [latex]{a}_{1}[\/latex], which is given as part of the formula. This is the first term.<\/li>\n \t<li>To find the second term, [latex]{a}_{2}[\/latex], substitute the initial term into the formula for [latex]{a}_{n - 1}[\/latex]. Solve.<\/li>\n \t<li>To find the third term, [latex]{a}_{3}[\/latex], substitute the second term into the formula. Solve.<\/li>\n \t<li>Repeat until you have solved for the [latex]n\\text{th}[\/latex] term.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Terms of a Sequence Defined by a Recursive Formula<\/h3>\nWrite the first five terms of the sequence defined by the recursive formula.\n<p style=\"text-align: center;\">[latex]\\begin{align} {a}_{1}&amp;=9 \\\\ {a}_{n}&amp;=3{a}_{n - 1}-20\\text{, for }n\\ge 2 \\end{align}[\/latex]<\/p>\n[reveal-answer q=\"748916\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"748916\"]\n\nThe first term is given in the formula. For each subsequent term, we replace [latex]{a}_{n - 1}[\/latex] with the value of the preceding term.\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;n=1 &amp;&amp; {a}_{1}=9 \\\\ &amp;n=2 &amp;&amp; {a}_{2}=3{a}_{1}-20=3\\left(9\\right)-20=27 - 20=7 \\\\ &amp;n=3 &amp;&amp; {a}_{3}=3{a}_{2}-20=3\\left(7\\right)-20=21 - 20=1 \\\\ &amp;n=4 &amp;&amp; {a}_{4}=3{a}_{3}-20=3\\left(1\\right)-20=3 - 20=-17 \\\\ &amp;n=5 &amp;&amp; {a}_{5}=3{a}_{4}-20=3\\left(-17\\right)-20=-51 - 20=-71 \\end{align}[\/latex]<\/p>\nThe first five terms are [latex]\\left\\{9,7,1,-17,-71\\right\\}[\/latex]\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nWrite the first five terms of the sequence defined by the recursive formula.\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&amp;=2\\\\ {a}_{n}&amp;=2{a}_{n - 1}+1\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\n[reveal-answer q=\"378600\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"378600\"]\n\n[latex]\\left\\{2, 5, 11, 23, 47\\right\\}[\/latex]\n\n[\/hidden-answer]\n<iframe id=\"mom21\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5812&amp;theme=oea&amp;iframe_resize_id=mom21\" width=\"100%\" height=\"450\"><\/iframe>\n\n<\/div>\n<div>\n<h2>Key Concepts<\/h2>\n<ul>\n \t<li>A sequence is a list of numbers, called terms, written in a specific order.<\/li>\n \t<li>Explicit formulas define each term of a sequence using the position of the term.<\/li>\n \t<li>An explicit formula for the [latex]n\\text{th}[\/latex] term of a sequence can be written by analyzing the pattern of several terms.<\/li>\n \t<li>Recursive formulas define each term of a sequence using previous terms.<\/li>\n \t<li>Recursive formulas must state the initial term, or terms, of a sequence.<\/li>\n \t<li>A set of terms can be written by using a recursive formula.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<strong>explicit formula<\/strong> a formula that defines each term of a sequence in terms of its position in the sequence\n\n<strong>finite sequence<\/strong> a function whose domain consists of a finite subset of the positive integers [latex]\\left\\{1,2,\\dots n\\right\\}[\/latex] for some positive integer [latex]n[\/latex]\n\n<strong>infinite sequence<\/strong> a function whose domain is the set of positive integers\n\n<strong>nth term of a sequence<\/strong> a formula for the general term of a sequence\n\n<strong>recursive formula<\/strong> a formula that defines each term of a sequence using previous term(s)\n\n<strong>sequence<\/strong> a function whose domain is a subset of the positive integers\n\n<strong>term<\/strong> a number in a sequence\n\n<\/div>\n\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul class=\"ul1\">\n<li class=\"li2\"><span class=\"s1\">Write the terms of a sequence defined by an explicit formula.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Write the terms of a sequence defined by a recursive formula.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Use factorial notation.<\/span><\/li>\n<\/ul>\n<\/div>\n<p>A video game company launches an exciting new advertising campaign. They predict the number of online visits to their website, or hits, will double each day. The model they are using shows 2 hits the first day, 4 hits the second day, 8 hits the third day, and so on.<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong>Day<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>\u2026<\/td>\n<\/tr>\n<tr>\n<td><strong>Hits<\/strong><\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<td>16<\/td>\n<td>32<\/td>\n<td>\u2026<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>If their model continues, how many hits will there be at the end of the month? To answer this question, we\u2019ll first need to know how to determine a list of numbers written in a specific order. In this section we will explore these kinds of ordered lists.<\/p>\n<h2>Sequences Defined by an Explicit Formula<\/h2>\n<p>One way to describe an ordered list of numbers is as a <strong>sequence<\/strong>. A sequence is a function whose domain is a subset of the counting numbers. The sequence established by the number of hits on the website is<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots \\right\\}[\/latex].<\/p>\n<p>The <strong>ellipsis<\/strong> (\u2026) indicates that the sequence continues indefinitely. Each number in the sequence is called a <strong>term<\/strong>. The first five terms of this sequence are 2, 4, 8, 16, and 32.<\/p>\n<p>Listing all of the terms for a sequence can be cumbersome. For example, finding the number of hits on the website at the end of the month would require listing out as many as 31 terms. A more efficient way to determine a specific term is by writing a formula to define the sequence.<\/p>\n<p>One type of formula is an <strong>explicit formula<\/strong>, which defines the terms of a sequence using their position in the sequence. Explicit formulas are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the <strong> [latex]n\\text{th}[\/latex] term of the sequence<\/strong>, where [latex]n[\/latex] is any positive number. In our example, each number in the sequence is double the previous number, so we can use powers of 2 to write a formula for the [latex]n\\text{th}[\/latex] term.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03220152\/CNX_Precalc_Figure_11_01_0012.jpg\" alt=\"Sequence of {2, 4, 8, 16, 32, ...} expressed in exponential form (i.e., {2^1, 2^2, 2^3, ..., 2^n, ...}\" width=\"487\" height=\"89\" \/><\/p>\n<p>The first term of the sequence is [latex]{2}^{1}=2[\/latex], the second term is [latex]{2}^{2}=4[\/latex], the third term is [latex]{2}^{3}=8[\/latex], and so on. The [latex]n\\text{th}[\/latex] term of the sequence can be found by raising 2 to the [latex]n\\text{th}[\/latex] power. An explicit formula for a sequence is named by a lower case letter [latex]a,b,c..[\/latex]. with the subscript [latex]n[\/latex]. The explicit formula for this sequence is<\/p>\n<p style=\"text-align: center;\">[latex]{a}_{n}={2}^{n}[\/latex]<\/p>\n<p>Now that we have a formula for the [latex]n\\text{th}[\/latex] term of the sequence, we can answer the question posed at the beginning of this section. We were asked to find the number of hits at the end of the month, which we will take to be 31 days. To find the number of hits on the last day of the month, we need to find the 31<sup>st<\/sup> term of the sequence. We will substitute 31 for [latex]n[\/latex] in the formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{31}&={2}^{31} \\\\ &=\\text{2,147,483,648} \\end{align}[\/latex]<\/p>\n<p>If the doubling trend continues, the company will get [latex]\\text{2,147,483,648}[\/latex] hits on the last day of the month. That is over 2.1 billion hits! The huge number is probably a little unrealistic because it does not take consumer interest and competition into account. It does, however, give the company a starting point from which to consider business decisions.<\/p>\n<p>Another way to represent the sequence is by using a table. The first five terms of the sequence and the [latex]n\\text{th}[\/latex] term of the sequence are shown in the table.<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong> [latex]n[\/latex] <\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>[latex]n[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]n\\text{th}[\/latex] term of the sequence, [latex]{a}_{n}[\/latex] <\/strong><\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<td>16<\/td>\n<td>32<\/td>\n<td>[latex]{2}^{n}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Graphing provides a visual representation of the sequence as a set of distinct points. We can see from the graph below that the number of hits is rising at an exponential rate. This particular sequence forms an exponential function.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03220155\/CNX_Precalc_Figure_11_01_0022.jpg\" alt=\"Graph of a plotted exponential function, f(n) = 2^n, where the x-axis is labeled n and the y-axis is labeled a_n.\" width=\"487\" height=\"397\" \/><\/p>\n<p>Lastly, we can write this particular sequence as<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots ,{2}^{n},\\dots \\right\\}[\/latex].<\/p>\n<p>A sequence that continues indefinitely is called an <strong>infinite sequence<\/strong>. The domain of an infinite sequence is the set of counting numbers. If we consider only the first 10 terms of the sequence, we could write<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots ,1024\\right\\}[\/latex].<\/p>\n<p>This sequence is called a <strong>finite sequence<\/strong> because it does not continue indefinitely.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Sequence<\/h3>\n<p>A <strong>sequence<\/strong> is a function whose domain is the set of positive integers. A <strong>finite sequence<\/strong> is a sequence whose domain consists of only the first [latex]n[\/latex] positive integers. The numbers in a sequence are called <strong>terms<\/strong>. The variable [latex]a[\/latex] with a number subscript is used to represent the terms in a sequence and to indicate the position of the term in the sequence.<\/p>\n<p style=\"text-align: center;\">[latex]{a}_{1},{a}_{2},{a}_{3},\\dots ,{a}_{n},\\dots[\/latex]<\/p>\n<p>We call [latex]{a}_{1}[\/latex] the first term of the sequence, [latex]{a}_{2}[\/latex] the second term of the sequence, [latex]{a}_{3}[\/latex] the third term of the sequence, and so on. The term [latex]{a}_{n}[\/latex] is called the <strong> [latex]n\\text{th}[\/latex] term of the sequence<\/strong>, or the general term of the sequence. An <strong>explicit formula<\/strong> defines the [latex]n\\text{th}[\/latex] term of a sequence using the position of the term. A sequence that continues indefinitely is an <strong>infinite sequence<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Does a sequence always have to begin with [latex]{a}_{1}?[\/latex]<\/h3>\n<p><em>No. In certain problems, it may be useful to define the initial term as [latex]{a}_{0}[\/latex] instead of [latex]{a}_{1}[\/latex]. In these problems, the domain of the function includes 0.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an explicit formula, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\n<ol>\n<li>Substitute each value of [latex]n[\/latex] into the formula. Begin with [latex]n=1[\/latex] to find the first term, [latex]{a}_{1}[\/latex].<\/li>\n<li>To find the second term, [latex]{a}_{2}[\/latex], use [latex]n=2[\/latex].<\/li>\n<li>Continue in the same manner until you have identified all [latex]n[\/latex] terms.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Terms of a Sequence Defined by an Explicit Formula<\/h3>\n<p>Write the first five terms of the sequence defined by the explicit formula [latex]{a}_{n}=-3n+8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q776421\">Show Solution<\/span><\/p>\n<div id=\"q776421\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]n=1[\/latex] into the formula. Repeat with values 2 through 5 for [latex]n[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{lign}n=1 && {a}_{1}=-3\\left(1\\right)+8=5 \\\\ n=2 && {a}_{2}=-3\\left(2\\right)+8=2 \\\\ n=3 && {a}_{3}=-3\\left(3\\right)+8=-1 \\\\ n=4 && {a}_{4}=-3\\left(4\\right)+8=-4 \\\\ n=5 && {a}_{5}=-3\\left(5\\right)+8=-7 \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">The first five terms are [latex]\\left\\{5,2,-1,-4,-7\\right\\}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The sequence values can be listed in a table. A table&nbsp;is a convenient way to input the function into a graphing utility.<\/p>\n<table summary=\"\">\n<tbody>\n<tr>\n<td><strong> [latex]n[\/latex] <\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td><strong> [latex]{a}_{n}[\/latex] <\/strong><\/td>\n<td>5<\/td>\n<td>2<\/td>\n<td>\u20131<\/td>\n<td>\u20134<\/td>\n<td>\u20137<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>A graph can be made from this table of values. From the graph in Figure 2, we can see that this sequence represents a linear function, but notice the graph is not continuous because the domain is over the positive integers only.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03220157\/CNX_Precalc_Figure_11_01_0032.jpg\" alt=\"Graph of a scattered plot where the x-axis is labeled n and the y-axis is labeled a_n.\" width=\"487\" height=\"584\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the first five terms of the sequence defined by the <strong>explicit formula<\/strong> [latex]{t}_{n}=5n - 4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q25802\">Show Solution<\/span><\/p>\n<div id=\"q25802\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first five terms are [latex]\\left\\{1,6, 11, 16, 21\\right\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5846&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Finding an Explicit Formula<\/h2>\n<p>Thus far, we have been given the explicit formula and asked to find a number of terms of the sequence. Sometimes, the explicit formula for the [latex]n\\text{th}[\/latex] term of a sequence is not given. Instead, we are given several terms from the sequence. When this happens, we can work in reverse to find an explicit formula from the first few terms of a sequence. The key to finding an explicit formula is to look for a pattern in the terms. Keep in mind that the pattern may involve alternating terms, formulas for numerators, formulas for denominators, exponents, or bases.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the first few terms of a sequence, find an explicit formula for the sequence.<\/h3>\n<ol>\n<li>Look for a pattern among the terms.<\/li>\n<li>If the terms are fractions, look for a separate pattern among the numerators and denominators.<\/li>\n<li>Look for a pattern among the signs of the terms.<\/li>\n<li>Write a formula for [latex]{a}_{n}[\/latex] in terms of [latex]n[\/latex]. Test your formula for [latex]n=1,n=2[\/latex],&nbsp;and&nbsp;[latex]n=3[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing an Explicit Formula for the <em style=\"font-size: 16px;\">n<\/em><span style=\"font-size: 16px;\">th Term of a Sequence<\/span><\/h3>\n<p>Write an explicit formula for the [latex]n\\text{th}[\/latex] term of each sequence.<\/p>\n<ol>\n<li>[latex]\\left\\{-\\dfrac{2}{11},\\dfrac{3}{13},-\\dfrac{4}{15},\\dfrac{5}{17},-\\dfrac{6}{19},\\dots \\right\\}[\/latex]<\/li>\n<li>[latex]\\left\\{-\\dfrac{2}{25},-\\dfrac{2}{125},-\\dfrac{2}{625},-\\dfrac{2}{3\\text{,}125},-\\dfrac{2}{15\\text{,}625},\\dots \\right\\}[\/latex]<\/li>\n<li>[latex]\\left\\{{e}^{4},{e}^{5},{e}^{6},{e}^{7},{e}^{8},\\dots \\right\\}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q328784\">Show Solution<\/span><\/p>\n<div id=\"q328784\" class=\"hidden-answer\" style=\"display: none\">\n<p>Look for the pattern in each sequence.<\/p>\n<ol>\n<li>The terms alternate between positive and negative. We can use [latex]{\\left(-1\\right)}^{n}[\/latex] to make the terms alternate. The numerator can be represented by [latex]n+1[\/latex]. The denominator can be represented by [latex]2n+9[\/latex].\n<div style=\"text-align: center;\">[latex]{a}_{n}=\\dfrac{{\\left(-1\\right)}^{n}\\left(n+1\\right)}{2n+9}[\/latex]<\/div>\n<\/li>\n<li>The terms are all negative.\n<div style=\"text-align: center;\">[latex]\\begin{align} & \\left\\{-\\dfrac{2}{25},-\\dfrac{2}{125},-\\dfrac{2}{625},-\\dfrac{2}{3\\text{,}125},-\\dfrac{2}{15\\text{,}625},\\dots \\right\\}&& \\text{The numerator is 2.} \\\\[1mm] & \\left\\{-\\dfrac{2}{5^2},-\\dfrac{2}{5^3},- \\dfrac{2}{5^4},-\\dfrac{2}{5^5},-\\dfrac{2}{5^6},\\dots,-\\dfrac{2}{5^n},\\dots \\right\\} && \\text{The denominators are increasing powers of 5.}\\end{align}[\/latex]<\/div>\n<p>So we know that the fraction is negative, the numerator is 2, and the denominator can be represented by [latex]{5}^{n+1}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]{a}_{n}=-\\dfrac{2}{{5}^{n+1}}[\/latex]<\/div>\n<\/li>\n<li>The terms are powers of [latex]e[\/latex]. For [latex]n=1[\/latex], the first term is [latex]{e}^{4}[\/latex] so the exponent must be [latex]n+3[\/latex].\n<div style=\"text-align: center;\">[latex]{a}_{n}={e}^{n+3}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence.<\/p>\n<p style=\"text-align: center;\">[latex]\\{9;\u221281,729;\u22126,561;59,049\\}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q806943\">Show Solution<\/span><\/p>\n<div id=\"q806943\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{a}_{n}={\\left(-1\\right)}^{n+1}{9}^{n}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5805&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"550\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence.<\/p>\n<div style=\"text-align: center;\">[latex]\\left\\{-\\dfrac{3}{4},-\\dfrac{9}{8},-\\dfrac{27}{12},-\\dfrac{81}{16},-\\dfrac{243}{20},\\dots\\right\\}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q282988\">Show Solution<\/span><\/p>\n<div id=\"q282988\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{a}_{n}=-\\dfrac{{3}^{n}}{4n}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5806&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"550\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence.<\/p>\n<div style=\"text-align: center;\">[latex]\\left\\{\\dfrac{1}{{e}^{2}}, \\dfrac{1}{e}, 1, e, {e}^{2},...\\right\\}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q804020\">Show Solution<\/span><\/p>\n<div id=\"q804020\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{a}_{n}={e}^{n - 3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Investigating Alternating Sequences<\/h2>\n<p>Sometimes sequences have terms that alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. Let\u2019s take a look at the following sequence.<\/p>\n<div style=\"text-align: center;\">[latex]\\left\\{2,-4,6,-8\\right\\}[\/latex]<\/div>\n<p>Notice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an explicit formula with alternating terms, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\n<ol>\n<li>Substitute each value of [latex]n[\/latex] into the formula. Begin with [latex]n=1[\/latex] to find the first term, [latex]{a}_{1}[\/latex]. The sign of the term in the explicit formula is given by the [latex]{\\left(-1\\right)}^{n}[\/latex] if the first term is negative and&nbsp;[latex]{\\left(-1\\right)}^{n-1}[\/latex] if the first term is positive.<\/li>\n<li>To find the second term, [latex]{a}_{2}[\/latex], use [latex]n=2[\/latex].<\/li>\n<li>Continue in the same manner until you have identified all [latex]n[\/latex] terms.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Terms of an Alternating Sequence Defined by an Explicit Formula<\/h3>\n<p>Write the first five terms of the sequence.<\/p>\n<div style=\"text-align: center;\">[latex]{a}_{n}=\\dfrac{{\\left(-1\\right)}^{n}{n}^{2}}{n+1}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q979322\">Show Solution<\/span><\/p>\n<div id=\"q979322\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]n=1[\/latex], [latex]n=2[\/latex], and so on in the formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&n=1 && {a}_{1}=\\dfrac{{\\left(-1\\right)}^{1}{2}^{2}}{1+1}=-\\dfrac{1}{2} \\\\[1mm] &n=2 && {a}_{2}=\\dfrac{{\\left(-1\\right)}^{2}{2}^{2}}{2+1}=\\dfrac{4}{3} \\\\[1mm] &n=3 && {a}_{3}=\\dfrac{{\\left(-1\\right)}^{3}{3}^{2}}{3+1}=-\\frac{9}{4} \\\\[1mm] &n=4 && {a}_{4}=\\dfrac{{\\left(-1\\right)}^{4}{4}^{2}}{4+1}=\\dfrac{16}{5} \\\\[1mm] &n=5 && {a}_{5}=\\dfrac{{\\left(-1\\right)}^{5}{5}^{2}}{5+1}=-\\dfrac{25}{6} \\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>The first five terms are [latex]\\left\\{-\\dfrac{1}{2},\\dfrac{4}{3},-\\dfrac{9}{4},\\dfrac{16}{5},-\\dfrac{25}{6}\\right\\}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The graph of this function looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03220159\/CNX_Precalc_Figure_11_01_0042.jpg\" alt=\"Graph of a scattered plot with labeled points: (1, -1\/2), (2, 4\/3), (3, -9\/4), (4, 16\/5), and (5, -25\/6). The x-axis is labeled n and the y-axis is labeled a_n.\" width=\"487\" height=\"482\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>In Example: Writing the Terms of an Alternating Sequence Defined by an Explicit Formula, does the (\u20131) to the power of [latex]n[\/latex] account for the oscillations of signs?<\/h4>\n<p><em>Yes, the power might be<\/em> [latex]n,n+1,n - 1[\/latex], <em>and so on, but any odd powers will result in a negative term, and any even power will result in a positive term.<\/em><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the first five terms of the sequence:<\/p>\n<div style=\"text-align: center;\">[latex]{a}_{n}=\\dfrac{4n}{{\\left(-2\\right)}^{n}}[\/latex]<\/div>\n<div><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q196840\">Show Solution<\/span><\/p>\n<div id=\"q196840\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first five terms are [latex]\\left\\{-2, 2, -\\dfrac{3}{2}, 1,-\\dfrac{5}{8}\\right\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom30\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5823&amp;theme=oea&amp;iframe_resize_id=mom30\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Sequences Defined by a Recursive Formula<\/h2>\n<p>Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34,\u2026. Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with 13 petals, and different varieties of daisies that may have 21 or 34 petals.<\/p>\n<p>Each term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a <strong>recursive formula<\/strong>, a formula that defines the terms of a sequence using previous terms.<\/p>\n<p>A recursive formula always has two parts: the value of an initial term (or terms), and an equation defining [latex]{a}_{n}[\/latex] in terms of preceding terms. For example, suppose we know the following:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&=3 \\\\ {a}_{n}&=2{a}_{n - 1}-1, \\text{for } n\\ge 2 \\end{align}[\/latex]<\/p>\n<p>We can find the subsequent terms of the sequence using the first term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&=3\\\\ {a}_{2}&=2{a}_{1}-1=2\\left(3\\right)-1=5\\\\ {a}_{3}&=2{a}_{2}-1=2\\left(5\\right)-1=9\\\\ {a}_{4}&=2{a}_{3}-1=2\\left(9\\right)-1=17\\end{align}[\/latex]<\/p>\n<p>So the first four terms of the sequence are [latex]\\left\\{3,5,9,17\\right\\}[\/latex].<\/p>\n<p>The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&=1 \\\\ {a}_{2}&=1 \\\\ {a}_{n}&={a}_{n - 1}+{a}_{n - 2}, \\text{for } n\\ge 3 \\end{align}[\/latex]<\/p>\n<p>To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We saw above that the eighth and ninth terms are 21 and 34, so<\/p>\n<p style=\"text-align: center;\">[latex]{a}_{10}={a}_{9}+{a}_{8}=34+21=55[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Recursive Formula<\/h3>\n<p>A <strong>recursive formula<\/strong> is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Must the first two terms always be given in a recursive formula?<\/h4>\n<p><em>No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only the first term to be defined.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a recursive formula with only the first term provided, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\n<ol>\n<li>Identify the initial term, [latex]{a}_{1}[\/latex], which is given as part of the formula. This is the first term.<\/li>\n<li>To find the second term, [latex]{a}_{2}[\/latex], substitute the initial term into the formula for [latex]{a}_{n - 1}[\/latex]. Solve.<\/li>\n<li>To find the third term, [latex]{a}_{3}[\/latex], substitute the second term into the formula. Solve.<\/li>\n<li>Repeat until you have solved for the [latex]n\\text{th}[\/latex] term.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Terms of a Sequence Defined by a Recursive Formula<\/h3>\n<p>Write the first five terms of the sequence defined by the recursive formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} {a}_{1}&=9 \\\\ {a}_{n}&=3{a}_{n - 1}-20\\text{, for }n\\ge 2 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q748916\">Show Solution<\/span><\/p>\n<div id=\"q748916\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first term is given in the formula. For each subsequent term, we replace [latex]{a}_{n - 1}[\/latex] with the value of the preceding term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&n=1 && {a}_{1}=9 \\\\ &n=2 && {a}_{2}=3{a}_{1}-20=3\\left(9\\right)-20=27 - 20=7 \\\\ &n=3 && {a}_{3}=3{a}_{2}-20=3\\left(7\\right)-20=21 - 20=1 \\\\ &n=4 && {a}_{4}=3{a}_{3}-20=3\\left(1\\right)-20=3 - 20=-17 \\\\ &n=5 && {a}_{5}=3{a}_{4}-20=3\\left(-17\\right)-20=-51 - 20=-71 \\end{align}[\/latex]<\/p>\n<p>The first five terms are [latex]\\left\\{9,7,1,-17,-71\\right\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the first five terms of the sequence defined by the recursive formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{a}_{1}&=2\\\\ {a}_{n}&=2{a}_{n - 1}+1\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q378600\">Show Solution<\/span><\/p>\n<div id=\"q378600\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left\\{2, 5, 11, 23, 47\\right\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom21\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5812&amp;theme=oea&amp;iframe_resize_id=mom21\" width=\"100%\" height=\"450\"><\/iframe><\/p>\n<\/div>\n<div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>A sequence is a list of numbers, called terms, written in a specific order.<\/li>\n<li>Explicit formulas define each term of a sequence using the position of the term.<\/li>\n<li>An explicit formula for the [latex]n\\text{th}[\/latex] term of a sequence can be written by analyzing the pattern of several terms.<\/li>\n<li>Recursive formulas define each term of a sequence using previous terms.<\/li>\n<li>Recursive formulas must state the initial term, or terms, of a sequence.<\/li>\n<li>A set of terms can be written by using a recursive formula.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<p><strong>explicit formula<\/strong> a formula that defines each term of a sequence in terms of its position in the sequence<\/p>\n<p><strong>finite sequence<\/strong> a function whose domain consists of a finite subset of the positive integers [latex]\\left\\{1,2,\\dots n\\right\\}[\/latex] for some positive integer [latex]n[\/latex]<\/p>\n<p><strong>infinite sequence<\/strong> a function whose domain is the set of positive integers<\/p>\n<p><strong>nth term of a sequence<\/strong> a formula for the general term of a sequence<\/p>\n<p><strong>recursive formula<\/strong> a formula that defines each term of a sequence using previous term(s)<\/p>\n<p><strong>sequence<\/strong> a function whose domain is a subset of the positive integers<\/p>\n<p><strong>term<\/strong> a number in a sequence<\/p>\n<\/div>\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-1856\">\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>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 5846, 5823. <strong>Authored by<\/strong>: Webwork-Rochester. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 5805, 5806, 5812. <strong>Authored by<\/strong>: David Lippman. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 6107. <strong>Authored by<\/strong>: Gregg Harbaugh. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 19458. <strong>Authored by<\/strong>: James Sousa. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License, CC-BY + GPL<\/li><li>Question ID 68773. <strong>Authored by<\/strong>: Roy Shahbazian. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":708740,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 5846, 5823\",\"author\":\"Webwork-Rochester\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 5805, 5806, 5812\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 6107\",\"author\":\"Gregg Harbaugh\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 19458\",\"author\":\"James Sousa\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License, CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 68773\",\"author\":\"Roy Shahbazian\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + 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