Algebra and Trigonometry 10th Edition

Published by Cengage Learning

Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 796: 56

Answer

$\displaystyle \sum_{n=1}^{10}(\frac{3}{2})^{n-1}=\frac{58025}{512}$

Work Step by Step

$\displaystyle \sum_{n=1}^{10}(\frac{3}{2})^{n-1}=(\frac{3}{2})^0+(\frac{3}{2})^1+...+(\frac{3}{2})^9$ $a_1=(\frac{3}{2})^0=1$ $r=\frac{a_2}{a_1}=\frac{(\frac{3}{2})^1}{(\frac{3}{2})^0}=\frac{3}{2}$ $S_{10}=a_1(\frac{1-r^n}{1-r})=1(\frac{1-(\frac{3}{2})^{10}}{1-(\frac{3}{2})})=\frac{1-(\frac{59049}{1024})}{-\frac{1}{2}}=\frac{-(\frac{58025}{1024})}{-\frac{1}{2}}=\frac{58025}{512}$

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