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: 11

Answer

$\Sigma_{k=0}^{9}(-\frac{3}{4})^k =.53925$

Work Step by Step

$\Sigma_{k=0}^{9}(-\frac{3}{4})^k =\frac{a(1-(r)^n)}{1-r}$ $a=(-\frac{3}{4})^0=1$ $r=-\frac{3}{4}$ $n= 10$ $$\Sigma_{k=0}^{9}(-\frac{3}{4})^k =\frac{1(1-(-\frac{3}{4})^{10})}{1-(-\frac{3}{4})}=.53925$$
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