{"id":14773,"date":"2018-09-27T18:35:26","date_gmt":"2018-09-27T18:35:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/sequences-and-their-notations-2\/"},"modified":"2021-06-30T15:34:28","modified_gmt":"2021-06-30T15:34:28","slug":"sequences-and-their-notations-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/nwfsc-MGF1107\/chapter\/sequences-and-their-notations-2\/","title":{"raw":"Sequences and Their Notations","rendered":"Sequences and Their Notations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Write the terms of a sequence defined by an explicit formula.<\/li>\r\n \t<li style=\"font-weight: 400;\">Write the terms of a sequence defined by a recursive formula.<\/li>\r\n \t<li>Write a recursive formula for a sequence.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Writing the Terms of a Sequence Defined by an Explicit Formula<\/h2>\r\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\r\n<div style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots \\right\\}[\/latex].<\/div>\r\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.\r\n\r\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.\r\n\r\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.\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\/3675\/2018\/09\/27183455\/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\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\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\r\n<div style=\"text-align: center;\">[latex]{a}_{n}={2}^{n}[\/latex].<\/div>\r\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.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{31}&amp;={2}^{31} \\\\ &amp;=\\text{2,147,483,648} \\end{align}[\/latex]<\/div>\r\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.\r\n\r\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.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]n[\/latex] <\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>[latex]n[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]n\\text{th}[\/latex] term of the sequence, [latex]{a}_{n}[\/latex] <\/strong><\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>8<\/td>\r\n<td>16<\/td>\r\n<td>32<\/td>\r\n<td>[latex]{2}^{n}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLastly, we can write this particular sequence as\r\n<div style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots ,{2}^{n},\\dots \\right\\}[\/latex].<\/div>\r\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\r\n<div style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots ,{2}^{n},\\dots ,1024\\right\\}[\/latex].<\/div>\r\nThis sequence is called a <strong>finite sequence<\/strong> because it does not continue indefinitely.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Sequence<\/h3>\r\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.\r\n<div style=\"text-align: center;\">[latex]{a}_{1},{a}_{2},{a}_{3},\\dots ,{a}_{n},\\dots [\/latex]<\/div>\r\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>.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an explicit formula, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\r\n<ol>\r\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>\r\n \t<li>To find the second term, [latex]{a}_{2}[\/latex], use [latex]n=2[\/latex].<\/li>\r\n \t<li>Continue in the same manner until you have identified all [latex]n[\/latex] terms.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Writing the Terms of a Sequence Defined by an Explicit Formula<\/h3>\r\nWrite the first five terms of the sequence defined by the explicit formula [latex]{a}_{n}=-3n+8[\/latex].\r\n\r\n[reveal-answer q=\"805645\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"805645\"]\r\n\r\nSubstitute [latex]n=1[\/latex] into the formula. Repeat with values 2 through 5 for [latex]n[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;n=1&amp;&amp; {a}_{1}=-3\\left(1\\right)+8=5 \\\\ &amp;n=2&amp;&amp; {a}_{2}=-3\\left(2\\right)+8=2 \\\\ &amp;n=3&amp;&amp; {a}_{3}=-3\\left(3\\right)+8=-1 \\\\ &amp;n=4&amp;&amp; {a}_{4}=-3\\left(4\\right)+8=-4\\\\ &amp;n=5&amp;&amp; {a}_{5}=-3\\left(5\\right)+8=-7 \\end{align}[\/latex]<\/p>\r\nThe first five terms are [latex]\\left\\{5,2,-1,-4,-7\\right\\}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe sequence values can be listed in a table.\r\n<table id=\"Table_11_01_03\" summary=\"\">\r\n<tbody>\r\n<tr>\r\n<td><strong> [latex]n[\/latex] <\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong> [latex]{a}_{n}[\/latex] <\/strong><\/td>\r\n<td>5<\/td>\r\n<td>2<\/td>\r\n<td>\u20131<\/td>\r\n<td>\u20134<\/td>\r\n<td>\u20137<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<span style=\"font-size: 1rem; text-align: initial;\">[\/hidden-answer]<\/span>\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the first five terms of the sequence defined by the <strong>explicit formula<\/strong> [latex]{t}_{n}=5n - 4[\/latex].\r\n\r\n[reveal-answer q=\"445186\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"445186\"]\r\n\r\nThe first five terms are [latex]\\left\\{1,6, 11, 16, 21\\right\\}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]172157[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Investigating Alternating Sequences<\/h2>\r\nSometimes sequences have terms that are alternate. In fact, the terms may actually alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. However, the resulting terms will not show increase or decrease as [latex]n[\/latex] increases. Let\u2019s take a look at the following sequence.\r\n<div style=\"text-align: center;\">[latex]\\left\\{2,-4,6,-8\\right\\}[\/latex]<\/div>\r\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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an explicit formula with alternating terms, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\r\n<ol>\r\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 is given by the [latex]{\\left(-1\\right)}^{n}[\/latex] in the explicit formula.<\/li>\r\n \t<li>To find the second term, [latex]{a}_{2}[\/latex], use [latex]n=2[\/latex].<\/li>\r\n \t<li>Continue in the same manner until you have identified all [latex]n[\/latex] terms.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Writing the Terms of an Alternating Sequence Defined by an Explicit Formula<\/h3>\r\nWrite the first five terms of the sequence.\r\n<p style=\"text-align: center;\">[latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n}{n}^{2}}{n+1}[\/latex]<\/p>\r\n[reveal-answer q=\"544526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"544526\"]\r\n\r\nSubstitute [latex]n=1[\/latex], [latex]n=2[\/latex], and so on in the formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;n=1&amp;&amp; {a}_{1}=\\frac{{\\left(-1\\right)}^{1}{2}^{2}}{1+1}=-\\frac{1}{2}\\\\ &amp;n=2&amp;&amp; {a}_{2}=\\frac{{\\left(-1\\right)}^{2}{2}^{2}}{2+1}=\\frac{4}{3}\\\\ &amp;n=3&amp;&amp; {a}_{3}=\\frac{{\\left(-1\\right)}^{3}{3}^{2}}{3+1}=-\\frac{9}{4} \\\\ &amp;n=4&amp;&amp; {a}_{4}=\\frac{{\\left(-1\\right)}^{4}{4}^{2}}{4+1}=\\frac{16}{5} \\\\ &amp;n=5&amp;&amp; {a}_{5}=\\frac{{\\left(-1\\right)}^{5}{5}^{2}}{5+1}=-\\frac{25}{6}\\end{align}[\/latex]<\/p>\r\nThe first five terms are [latex]\\left\\{-\\frac{1}{2},\\frac{4}{3},-\\frac{9}{4},\\frac{16}{5},-\\frac{25}{6}\\right\\}[\/latex].\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">[\/hidden-answer]<\/span>\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the first five terms of the sequence:\r\n<p style=\"text-align: center;\">[latex]{a}_{n}=\\frac{4n}{{\\left(-2\\right)}^{n}}[\/latex]<\/p>\r\n[reveal-answer q=\"93529\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"93529\"]\r\n\r\nThe first five terms are [latex]\\left\\{-2, 2, -\\frac{3}{2}, 1,\\text{ }-\\frac{5}{8}\\right\\}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Writing the Terms of a Sequence Defined by a Recursive Formula<\/h2>\r\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.\r\n\r\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.\r\n\r\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:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{a}_{1}=3 \\\\ &amp;{a}_{n}=2{a}_{n - 1}-1, \\text{ for } n\\ge 2 \\end{align}[\/latex]<\/div>\r\nWe can find the subsequent terms of the sequence using the first term.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{a}_{1}=3\\\\ &amp;{a}_{2}=2{a}_{1}-1=2\\left(3\\right)-1=5\\\\ &amp;{a}_{3}=2{a}_{2}-1=2\\left(5\\right)-1=9\\\\ &amp;{a}_{4}=2{a}_{3}-1=2\\left(9\\right)-1=17\\end{align}[\/latex]<\/div>\r\nSo the first four terms of the sequence are [latex]\\left\\{3,\\text{ }5,\\text{ }9,\\text{ }17\\right\\}[\/latex] .\r\n\r\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.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{a}_{1}=1 \\\\ &amp;{a}_{2}=1 \\\\ &amp;{a}_{n}={a}_{n - 1}+{a}_{n - 2}, \\text{ for } n\\ge 3 \\end{align}[\/latex]<\/div>\r\nTo find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are 21 and 34, so\r\n<div style=\"text-align: center;\">[latex]{a}_{10}={a}_{9}+{a}_{8}=34+21=55[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Recursive Formula<\/h3>\r\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.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\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>\r\n<ol>\r\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>\r\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>\r\n \t<li>To find the third term, [latex]{a}_{3}[\/latex], substitute the second term into the formula. Solve.<\/li>\r\n \t<li>Repeat until you have solved for the [latex]n\\text{th}[\/latex] term.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Writing the Terms of a Sequence Defined by a Recursive Formula<\/h3>\r\nWrite the first five terms of the sequence defined by the recursive formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{a}_{1}=9\\\\ &amp;{a}_{n}=3{a}_{n - 1}-20\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"638659\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"638659\"]\r\n\r\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.\r\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>\r\nThe first five terms are [latex]\\left\\{9,\\text{ }7,\\text{ }1,\\text{ }-17,\\text{ }-71\\right\\}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the first five terms of the sequence defined by the recursive formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{a}_{1}=2\\\\ &amp;{a}_{n}=2{a}_{n - 1}+1\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"961830\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"961830\"]\r\n\r\n[latex]\\left\\{2, 5, 11, 23, 47\\right\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]172190[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a recursive formula with two initial terms, write the first [latex]n[\/latex] terms of a sequence.<\/h3>\r\n<ol>\r\n \t<li>Identify the initial term, [latex]{a}_{1}[\/latex], which is given as part of the formula.<\/li>\r\n \t<li>Identify the second term, [latex]{a}_{2}[\/latex], which is given as part of the formula.<\/li>\r\n \t<li>To find the third term, substitute the initial term and the second term into the formula. Evaluate.<\/li>\r\n \t<li>Repeat until you have evaluated the [latex]n\\text{th}[\/latex] term.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Writing the Terms of a Sequence Defined by a Recursive Formula<\/h3>\r\nWrite the first six terms of the sequence defined by the recursive formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{a}_{1}=1\\\\ &amp;{a}_{2}=2\\\\ &amp;{a}_{n}=3{a}_{n - 1}+4{a}_{n - 2}\\text{, for }n\\ge 3\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"766687\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"766687\"]\r\n\r\nThe first two terms are given. For each subsequent term, we replace [latex]{a}_{n - 1}[\/latex] and [latex]{a}_{n - 2}[\/latex] with the values of the two preceding terms.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;n=3&amp;&amp; {a}_{3}=3{a}_{2}+4{a}_{1}=3\\left(2\\right)+4\\left(1\\right)=10 \\\\ &amp;n=4&amp;&amp; {a}_{4}=3{a}_{3}+4{a}_{2}=3\\left(10\\right)+4\\left(2\\right)=38 \\\\ &amp;n=5&amp;&amp; {a}_{5}=3{a}_{4}+4{a}_{3}=3\\left(38\\right)+4\\left(10\\right)=154 \\\\ &amp;n=6&amp;&amp; {a}_{6}=3{a}_{5}+4{a}_{4}=3\\left(154\\right)+4\\left(38\\right)=614 \\end{align}[\/latex]<\/p>\r\nThe first six terms are [latex]\\left\\{\\text{1, 2, 10, 38, 154, 614}\\right\\}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the first eight terms of the sequence defined by the recursive formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;{a}_{1}=0\\\\ &amp;{a}_{2}=1 \\\\ &amp;{a}_{3}=1 \\\\ &amp;{a}_{n}=\\frac{{a}_{n - 1}}{{a}_{n - 2}}+{a}_{n - 3}\\text{, for }n\\ge 4 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"731446\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"731446\"]\r\n\r\n[latex]\\left\\{0, 1, 1, 1, 2, 3, \\frac{5}{2},\\text{ }\\frac{17}{6}\\right\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5812[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using Factorial Notation<\/h2>\r\nThe formulas for some sequences include products of consecutive positive integers. <strong> [latex]n[\/latex] factorial<\/strong>, written as [latex]n![\/latex], is the product of the positive integers from 1 to [latex]n[\/latex]. For example,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;4!=4\\cdot 3\\cdot 2\\cdot 1=24 \\\\ &amp;5!=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120 \\end{align}[\/latex]<\/div>\r\nAn example of formula containing a <strong>factorial<\/strong> is [latex]{a}_{n}=\\left(n+1\\right)![\/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[\/latex].\r\n<div style=\"text-align: center;\">[latex]{a}_{6}=\\left(6+1\\right)!=7!=7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=5040[\/latex]<\/div>\r\nThe factorial of any whole number [latex]n[\/latex] is [latex]n\\left(n - 1\\right)![\/latex] We can therefore also think of [latex]5![\/latex] as [latex]5\\cdot 4!\\text{.}[\/latex]\r\n<div class=\"textbox\">\r\n<h3>A General Note: Factorial<\/h3>\r\n<strong><em>n<\/em> factorial<\/strong> is a mathematical operation that can be defined using a recursive formula. The factorial of [latex]n[\/latex], denoted [latex]n![\/latex], is defined for a positive integer [latex]n[\/latex] as:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;0!=1\\\\ &amp;1!=1\\\\ &amp;n!=n\\left(n - 1\\right)\\left(n - 2\\right)\\cdots \\left(2\\right)\\left(1\\right)\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\r\nThe special case [latex]0![\/latex] is defined as [latex]0!=1[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Writing the Terms of a Sequence Using Factorials<\/h3>\r\nWrite the first five terms of the sequence defined by the explicit formula [latex]{a}_{n}=\\frac{5n}{\\left(n+2\\right)!}[\/latex].\r\n\r\n[reveal-answer q=\"421735\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421735\"]\r\n\r\nSubstitute [latex]n=1,n=2[\/latex], and so on in the formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;n=1&amp;&amp; {a}_{1}=\\frac{5\\left(1\\right)}{\\left(1+2\\right)!}=\\frac{5}{3!}=\\frac{5}{3\\cdot 2\\cdot 1}=\\frac{5}{6} \\\\ &amp;n=2&amp;&amp; {a}_{2}=\\frac{5\\left(2\\right)}{\\left(2+2\\right)!}=\\frac{10}{4!}=\\frac{10}{4\\cdot 3\\cdot 2\\cdot 1}=\\frac{5}{12} \\\\ &amp;n=3&amp;&amp; {a}_{3}=\\frac{5\\left(3\\right)}{\\left(3+2\\right)!}=\\frac{15}{5!}=\\frac{15}{5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}=\\frac{1}{8} \\\\ &amp;n=4&amp;&amp; {a}_{4}=\\frac{5\\left(4\\right)}{\\left(4+2\\right)!}=\\frac{20}{6!}=\\frac{20}{6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}=\\frac{1}{36} \\\\ &amp;n=5&amp;&amp; {a}_{5}=\\frac{5\\left(5\\right)}{\\left(5+2\\right)!}=\\frac{25}{7!}=\\frac{25}{7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}=\\frac{5}{1\\text{,}008} \\end{align}[\/latex]<\/p>\r\nThe first five terms are [latex]\\left\\{\\frac{5}{6},\\frac{5}{12},\\frac{1}{8},\\frac{1}{36},\\frac{5}{1,008}\\right\\}[\/latex].\r\n\r\n[\/hidden-answer]<b><\/b>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the first five terms of the sequence defined by the explicit formula [latex]{a}_{n}=\\frac{\\left(n+1\\right)!}{2n}[\/latex].\r\n\r\n[reveal-answer q=\"179210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"179210\"]\r\n\r\nThe first five terms are [latex]\\left\\{1, \\frac{3}{2}, 4,\\text{ }15,\\text{ }72\\right\\}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]97805[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li style=\"font-weight: 400;\">Write the terms of a sequence defined by an explicit formula.<\/li>\n<li style=\"font-weight: 400;\">Write the terms of a sequence defined by a recursive formula.<\/li>\n<li>Write a recursive formula for a sequence.<\/li>\n<\/ul>\n<\/div>\n<h2>Writing the Terms of a Sequence 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<div style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots \\right\\}[\/latex].<\/div>\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<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183455\/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 class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\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<div style=\"text-align: center;\">[latex]{a}_{n}={2}^{n}[\/latex].<\/div>\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<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{31}&={2}^{31} \\\\ &=\\text{2,147,483,648} \\end{align}[\/latex]<\/div>\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>Lastly, we can write this particular sequence as<\/p>\n<div style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots ,{2}^{n},\\dots \\right\\}[\/latex].<\/div>\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<div style=\"text-align: center;\">[latex]\\left\\{2,4,8,16,32,\\dots ,{2}^{n},\\dots ,1024\\right\\}[\/latex].<\/div>\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<div style=\"text-align: center;\">[latex]{a}_{1},{a}_{2},{a}_{3},\\dots ,{a}_{n},\\dots[\/latex]<\/div>\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>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 shaded\">\n<h3>Example 1: 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=\"q805645\">Show Solution<\/span><\/p>\n<div id=\"q805645\" 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{align}&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>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.<\/p>\n<table id=\"Table_11_01_03\" 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><span style=\"font-size: 1rem; text-align: initial;\"><\/div>\n<\/div>\n<p><\/span><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\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=\"q445186\">Show Solution<\/span><\/p>\n<div id=\"q445186\" 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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm172157\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=172157&theme=oea&iframe_resize_id=ohm172157\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Investigating Alternating Sequences<\/h2>\n<p>Sometimes sequences have terms that are alternate. In fact, the terms may actually alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. However, the resulting terms will not show increase or decrease as [latex]n[\/latex] increases. 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 is given by the [latex]{\\left(-1\\right)}^{n}[\/latex] in the explicit formula.<\/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 shaded\">\n<h3>Example 2: 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<p style=\"text-align: center;\">[latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n}{n}^{2}}{n+1}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q544526\">Show Solution<\/span><\/p>\n<div id=\"q544526\" 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<p style=\"text-align: center;\">[latex]\\begin{align}&n=1&& {a}_{1}=\\frac{{\\left(-1\\right)}^{1}{2}^{2}}{1+1}=-\\frac{1}{2}\\\\ &n=2&& {a}_{2}=\\frac{{\\left(-1\\right)}^{2}{2}^{2}}{2+1}=\\frac{4}{3}\\\\ &n=3&& {a}_{3}=\\frac{{\\left(-1\\right)}^{3}{3}^{2}}{3+1}=-\\frac{9}{4} \\\\ &n=4&& {a}_{4}=\\frac{{\\left(-1\\right)}^{4}{4}^{2}}{4+1}=\\frac{16}{5} \\\\ &n=5&& {a}_{5}=\\frac{{\\left(-1\\right)}^{5}{5}^{2}}{5+1}=-\\frac{25}{6}\\end{align}[\/latex]<\/p>\n<p>The first five terms are [latex]\\left\\{-\\frac{1}{2},\\frac{4}{3},-\\frac{9}{4},\\frac{16}{5},-\\frac{25}{6}\\right\\}[\/latex].<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\"><\/div>\n<\/div>\n<p><\/span><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write the first five terms of the sequence:<\/p>\n<p style=\"text-align: center;\">[latex]{a}_{n}=\\frac{4n}{{\\left(-2\\right)}^{n}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q93529\">Show Solution<\/span><\/p>\n<div id=\"q93529\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first five terms are [latex]\\left\\{-2, 2, -\\frac{3}{2}, 1,\\text{ }-\\frac{5}{8}\\right\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Writing the Terms of a Sequence 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<div 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]<\/div>\n<p>We can find the subsequent terms of the sequence using the first term.<\/p>\n<div 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]<\/div>\n<p>So the first four terms of the sequence are [latex]\\left\\{3,\\text{ }5,\\text{ }9,\\text{ }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<div 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]<\/div>\n<p>To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are 21 and 34, so<\/p>\n<div style=\"text-align: center;\">[latex]{a}_{10}={a}_{9}+{a}_{8}=34+21=55[\/latex]<\/div>\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>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 shaded\">\n<h3>Example 5: 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=\"q638659\">Show Solution<\/span><\/p>\n<div id=\"q638659\" 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,\\text{ }7,\\text{ }1,\\text{ }-17,\\text{ }-71\\right\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\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=\"q961830\">Show Solution<\/span><\/p>\n<div id=\"q961830\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left\\{2, 5, 11, 23, 47\\right\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm172190\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=172190&theme=oea&iframe_resize_id=ohm172190\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a recursive formula with two initial terms, 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.<\/li>\n<li>Identify the second term, [latex]{a}_{2}[\/latex], which is given as part of the formula.<\/li>\n<li>To find the third term, substitute the initial term and the second term into the formula. Evaluate.<\/li>\n<li>Repeat until you have evaluated the [latex]n\\text{th}[\/latex] term.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Writing the Terms of a Sequence Defined by a Recursive Formula<\/h3>\n<p>Write the first six terms of the sequence defined by the recursive formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{a}_{1}=1\\\\ &{a}_{2}=2\\\\ &{a}_{n}=3{a}_{n - 1}+4{a}_{n - 2}\\text{, for }n\\ge 3\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q766687\">Show Solution<\/span><\/p>\n<div id=\"q766687\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first two terms are given. For each subsequent term, we replace [latex]{a}_{n - 1}[\/latex] and [latex]{a}_{n - 2}[\/latex] with the values of the two preceding terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&n=3&& {a}_{3}=3{a}_{2}+4{a}_{1}=3\\left(2\\right)+4\\left(1\\right)=10 \\\\ &n=4&& {a}_{4}=3{a}_{3}+4{a}_{2}=3\\left(10\\right)+4\\left(2\\right)=38 \\\\ &n=5&& {a}_{5}=3{a}_{4}+4{a}_{3}=3\\left(38\\right)+4\\left(10\\right)=154 \\\\ &n=6&& {a}_{6}=3{a}_{5}+4{a}_{4}=3\\left(154\\right)+4\\left(38\\right)=614 \\end{align}[\/latex]<\/p>\n<p>The first six terms are [latex]\\left\\{\\text{1, 2, 10, 38, 154, 614}\\right\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write the first eight terms of the sequence defined by the recursive formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &{a}_{1}=0\\\\ &{a}_{2}=1 \\\\ &{a}_{3}=1 \\\\ &{a}_{n}=\\frac{{a}_{n - 1}}{{a}_{n - 2}}+{a}_{n - 3}\\text{, for }n\\ge 4 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q731446\">Show Solution<\/span><\/p>\n<div id=\"q731446\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left\\{0, 1, 1, 1, 2, 3, \\frac{5}{2},\\text{ }\\frac{17}{6}\\right\\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5812\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5812&theme=oea&iframe_resize_id=ohm5812&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using Factorial Notation<\/h2>\n<p>The formulas for some sequences include products of consecutive positive integers. <strong> [latex]n[\/latex] factorial<\/strong>, written as [latex]n![\/latex], is the product of the positive integers from 1 to [latex]n[\/latex]. For example,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&4!=4\\cdot 3\\cdot 2\\cdot 1=24 \\\\ &5!=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120 \\end{align}[\/latex]<\/div>\n<p>An example of formula containing a <strong>factorial<\/strong> is [latex]{a}_{n}=\\left(n+1\\right)![\/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]{a}_{6}=\\left(6+1\\right)!=7!=7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=5040[\/latex]<\/div>\n<p>The factorial of any whole number [latex]n[\/latex] is [latex]n\\left(n - 1\\right)![\/latex] We can therefore also think of [latex]5![\/latex] as [latex]5\\cdot 4!\\text{.}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Factorial<\/h3>\n<p><strong><em>n<\/em> factorial<\/strong> is a mathematical operation that can be defined using a recursive formula. The factorial of [latex]n[\/latex], denoted [latex]n![\/latex], is defined for a positive integer [latex]n[\/latex] as:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&0!=1\\\\ &1!=1\\\\ &n!=n\\left(n - 1\\right)\\left(n - 2\\right)\\cdots \\left(2\\right)\\left(1\\right)\\text{, for }n\\ge 2\\end{align}[\/latex]<\/p>\n<p>The special case [latex]0![\/latex] is defined as [latex]0!=1[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Writing the Terms of a Sequence Using Factorials<\/h3>\n<p>Write the first five terms of the sequence defined by the explicit formula [latex]{a}_{n}=\\frac{5n}{\\left(n+2\\right)!}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q421735\">Show Solution<\/span><\/p>\n<div id=\"q421735\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]n=1,n=2[\/latex], and so on in the formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&n=1&& {a}_{1}=\\frac{5\\left(1\\right)}{\\left(1+2\\right)!}=\\frac{5}{3!}=\\frac{5}{3\\cdot 2\\cdot 1}=\\frac{5}{6} \\\\ &n=2&& {a}_{2}=\\frac{5\\left(2\\right)}{\\left(2+2\\right)!}=\\frac{10}{4!}=\\frac{10}{4\\cdot 3\\cdot 2\\cdot 1}=\\frac{5}{12} \\\\ &n=3&& {a}_{3}=\\frac{5\\left(3\\right)}{\\left(3+2\\right)!}=\\frac{15}{5!}=\\frac{15}{5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}=\\frac{1}{8} \\\\ &n=4&& {a}_{4}=\\frac{5\\left(4\\right)}{\\left(4+2\\right)!}=\\frac{20}{6!}=\\frac{20}{6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}=\\frac{1}{36} \\\\ &n=5&& {a}_{5}=\\frac{5\\left(5\\right)}{\\left(5+2\\right)!}=\\frac{25}{7!}=\\frac{25}{7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}=\\frac{5}{1\\text{,}008} \\end{align}[\/latex]<\/p>\n<p>The first five terms are [latex]\\left\\{\\frac{5}{6},\\frac{5}{12},\\frac{1}{8},\\frac{1}{36},\\frac{5}{1,008}\\right\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write the first five terms of the sequence defined by the explicit formula [latex]{a}_{n}=\\frac{\\left(n+1\\right)!}{2n}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q179210\">Show Solution<\/span><\/p>\n<div id=\"q179210\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first five terms are [latex]\\left\\{1, \\frac{3}{2}, 4,\\text{ }15,\\text{ }72\\right\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm97805\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=97805&theme=oea&iframe_resize_id=ohm97805&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/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-14773\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><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":17533,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax 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