Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 38



Work Step by Step

$f(x)=e^{3x}-e^{2x}$ Since, $e^{x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}$ $e^{3x}-e^{2x}=\Sigma_{n=0}^{\infty}\frac{(3x)^{n}}{n!}-\Sigma_{n=0}^{\infty}\frac{(2x)^{n}}{n!}$ $=\Sigma_{n=0}^{\infty}\frac{(3^{n}-2^{n})x^{n}}{n!}$
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