{"id":1682,"date":"2025-07-25T19:11:40","date_gmt":"2025-07-25T19:11:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1682"},"modified":"2026-03-24T22:30:18","modified_gmt":"2026-03-24T22:30:18","slug":"sequences-and-their-notations-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/sequences-and-their-notations-learn-it-2\/","title":{"raw":"Sequences and Their Notations: Learn It 2","rendered":"Sequences and Their Notations: Learn It 2"},"content":{"raw":"

Finding an Explicit Formula<\/h2>\r\nWhat if you\u2019re given a sequence and need to figure out the explicit formula for it?\r\n\r\nIn this case, you can work backward to discover the formula using the first few terms. The key is to look for patterns in the sequence. These patterns might involve alternating terms, or there could be specific formulas for the numerators, denominators, exponents, or bases. Once you spot the pattern, you can use it to create the explicit formula for the entire sequence.\r\n\r\n
Write an explicit formula for the [latex]n\\text{th}[\/latex] term of each sequence.\r\n
    \r\n \t
  1. [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>\r\n \t
  2. [latex]\\left\\{{e}^{4},{e}^{5},{e}^{6},{e}^{7},{e}^{8},\\dots \\right\\}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"328784\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"328784\"]\r\n\r\nLook for the pattern in each sequence.\r\n
      \r\n \t
    1. The terms are all negative.\r\n
      [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},-\r\n\\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>\r\nSo we know that the fraction is negative, the numerator is [latex]2[\/latex], and the denominator can be represented by [latex]{5}^{n+1}[\/latex].\r\n
      [latex]{a}_{n}=-\\dfrac{2}{{5}^{n+1}}[\/latex]<\/div><\/li>\r\n \t
    2. 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].\r\n
      [latex]{a}_{n}={e}^{n+3}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      [ohm_question hide_question_numbers=1]321843[\/ohm_question]<\/section>
      [ohm_question hide_question_numbers=1]321844[\/ohm_question]<\/section>\r\n

      Alternating Sequences<\/h3>\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.\r\n\r\nLet\u2019s take a look at the following sequence.\r\n
      [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\r\n
      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 [latex]{\\left(-1\\right)}^{n-1}[\/latex] if the first term is positive.<\/section>
      Write the first five terms of the sequence.\r\n
      [latex]{a}_{n}=\\dfrac{{\\left(-1\\right)}^{n}{n}^{2}}{n+1}[\/latex]<\/div>\r\n[reveal-answer q=\"979322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979322\"]\r\n\r\nSubstitute [latex]n=1[\/latex], [latex]n=2[\/latex], and so on in the formula.\r\n
      [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>\r\n
      <\/div>\r\nThe first five terms are [latex]\\left\\{-\\dfrac{1}{2},\\dfrac{4}{3},-\\dfrac{9}{4},\\dfrac{16}{5},-\\dfrac{25}{6}\\right\\}[\/latex].\r\n\r\nAnalysis of the Solution<\/strong>\r\n\r\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.\r\n\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
      [ohm_question hide_question_numbers=1]321845[\/ohm_question]<\/section>
      Write an explicit formula for the [latex]n\\text{th}[\/latex] term of each sequence.\r\n

      [latex]\\left\\{-\\dfrac{2}{11},\\dfrac{3}{13},-\\dfrac{4}{15},\\dfrac{5}{17},-\\dfrac{6}{19},\\dots \\right\\}[\/latex]<\/p>\r\n[reveal-answer q=\"560661\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"560661\"]\r\n\r\nThe 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].\r\n\r\nThe denominator can be represented by [latex]2n+9[\/latex].\r\n

      [latex]{a}_{n}=\\dfrac{{\\left(-1\\right)}^{n}\\left(n+1\\right)}{2n+9}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      [ohm_question hide_question_numbers=1]321846[\/ohm_question]<\/section>
      [ohm_question]321847[\/ohm_question]<\/section><\/section>","rendered":"

      Finding an Explicit Formula<\/h2>\n

      What if you\u2019re given a sequence and need to figure out the explicit formula for it?<\/p>\n

      In this case, you can work backward to discover the formula using the first few terms. The key is to look for patterns in the sequence. These patterns might involve alternating terms, or there could be specific formulas for the numerators, denominators, exponents, or bases. Once you spot the pattern, you can use it to create the explicit formula for the entire sequence.<\/p>\n

      Write an explicit formula for the [latex]n\\text{th}[\/latex] term of each sequence.<\/p>\n
        \n
      1. [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
      2. [latex]\\left\\{{e}^{4},{e}^{5},{e}^{6},{e}^{7},{e}^{8},\\dots \\right\\}[\/latex]<\/li>\n<\/ol>\n