Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.3 Infinite Series - 8.3 Exercises - Page 623: 12

Answer

$−70.46875$

Work Step by Step

We are given the geometric sum: $\sum_{k=1}^5 (-2.5)^k$ Rewrite the sum: $n=5$ $S_5=a+ar+...+ar^4=\sum_{k=0}^4 ar^k=a\dfrac{1-r^5}{1-r}$ The geometric sequence has the ratio $r=-2.5$ and the first term $a=-2.5$. $S_5=\sum_{k=0}^4 (-2.5)(-2.5)^k=(-2.5)\dfrac{1-(-2.5)^5}{1-(-2.5)}=−70.46875$
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