Algebra and Trigonometry 10th Edition

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

Chapter 11 - Review Exercises - Page 841: 54


$\displaystyle \sum_{i=1}^{6}(\frac{1}{3})^{i-1}=\frac{364}{243}$

Work Step by Step

$\displaystyle \sum_{i=1}^{6}(\frac{1}{3})^{i-1}=(\frac{1}{3})^{0}+(\frac{1}{3})^{1}+..+(\frac{1}{3})^{5}$ $a_1=(\frac{1}{3})^0=1$ $r=\frac{a_2}{a_1}=\frac{(\frac{1}{3})^{1}}{(\frac{1}{3})^{0}}=\frac{1}{3}$ $S_n=a_1(\frac{1-r^n}{1-r})$ $S_6=1(\frac{1-(\frac{1}{3})^6}{1-\frac{1}{3}})=\frac{1-\frac{1}{729}}{\frac{2}{3}}=\frac{\frac{728}{729}}{\frac{2}{3}}=\frac{364}{243}$
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