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Unformatted text preview: Solution : In order to do this, we need to gure out how many terms are necessary to add. Recall that X k = n +1 f ( k ) < Z n f ( x ) dx. Thus, we want to nd an integer n such that Z n (2 x + 1)6 dx < . 00001 . Z n (2 x + 1)6 dx = lim b Z b n (2 x + 1)6 dx = lim b 1 2 Z 2 b +1 2 n +1 u6 du = 1 2 lim b 1 5 u 5 2 b +1 2 n +1 = 1 2 lim b 1 5(2 b + 1) 51 5(2 n + 1) 5 = 1 10(2 n + 1) 5 . Thus, we want to nd an n for which 1 10(2 n +1) 5 < 1 100000 . Taking the reciprocal gives 10(2 n + 1) 5 > 100000, or (2 n + 1) 5 > 10000. Thus, we want n > 5 100001 2 2 . 65 . Thus, the sum of the rst n = 3 terms of the series, 1 3 6 + 1 5 6 + 1 7 6 . 001444242 is an approximation to the series that is correct to ve decimal places....
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 Spring '08
 REMPE
 Geometric Series

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