Stitz-Zeager_College_Algebra_e-book

# To generate the numbers in the middle of the rows

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Unformatted text preview: quence of numerators and denominators, respectively, we have an = dn . After n some experimentation,5 we choose to write the ﬁrst term as a fraction and associate the 4 negatives ‘−’ with the numerators. This yields 1 , −2 , 13 , −8 , . . .. The numerators form the 17 19 sequence 1, −2, 4, −8, . . . which is geometric with a = 1 and r = −2, so we get cn = (−2)n−1 , for n ≥ 1. The denominators 1, 7, 13, 19, . . . form an arithmetic sequence with a = 1 and d = 6. Hence, we get dn = 1 + 6(n − 1) = 6n − 5, for n ≥ 1. We obtain our formula for n−1 c an = dn = (−2)−5 , for n ≥ 1. We leave it to the reader to show that this checks out. 6n n While the last problem in Example 9.1.3 was neither geometric nor arithmetic, it did resolve into a combination of these two kinds of sequences. If handed the sequence 2, 5, 10, 17, . . ., we would be hard-pressed to ﬁnd a formula for an if we restrict our attention to these two archetypes. We said before that there is no general algorithm for ﬁnding the explicit formula for th...
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