Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 796: 63


$\displaystyle \sum_{j=0}^{40}2(-\frac{1}{4})^n\approx1.6$

Work Step by Step

$\displaystyle \sum_{j=0}^{40}2(-\frac{1}{4})^n=2(-\frac{1}{4})^0+2(-\frac{1}{4})^1+2(-\frac{1}{4})^2+...+2(-\frac{1}{4})^{40}$ There are 41 terms is this sequence. $a_1=2(-\frac{1}{4})^0=2$ $r=\frac{a_2}{a_1}=\frac{2(-\frac{1}{4})^1}{2(-\frac{1}{4})^0}=-\frac{1}{4}$ $S_n=a_1(\frac{1-r^n}{1-r})$ $S_{41}=2(\frac{1-(-\frac{1}{4})^{41}}{1-(-\frac{1}{4})})=2(\frac{1-(-\frac{1}{4})^{41}}{\frac{5}{4}})\approx1.6$
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