Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 38

Answer

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

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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