Stitz-Zeager_College_Algebra_e-book

Back then you were working in a base ten system here

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Unformatted text preview: rmine which value of n produces the term 117. Setting an = 117, we get 2n − 1 = 117 or n = 59. Our final answer is 59 1 + 3 + 5 + . . . + 117 = (2n − 1) n=1 (b) We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives ‘−’ with the numerators to get 1 −1 1 −1 1 + ++ + ... + 1 2 3 4 117 The numerators, 1, −1, etc. can be described by the geometric sequence1 cn = (−1)n−1 for n ≥ 1, while the denominators are given by the arithmetic sequence2 dn = n for n−1 n ≥ 1. Hence, we get the formula an = (−1) for our terms, and we find the lower and n upper limits of summation to be n = 1 and n = 117, respectively. Thus 111 1 1 − + − + −... + 234 117 117 = n=1 (−1)n−1 n 9 (c) Thanks to Example 9.1.3, we know that one formula for the nth term is an = 10n for n ≥ 1. This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the n − 1 zeros to the right of the decimal point before the 9, we...
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