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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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