University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 20


$2x^2-\dfrac{2^2x^3}{2}+\dfrac{2^3x^{4}}{3}-$ or, $\Sigma_{n=0}^\infty (-1)^n \dfrac{2^{n+1} x^{n+2}}{n+1}$

Work Step by Step

Since, we know that the Maclaurin Series for $ln(1+x)$ is defined as: $ \ln (1+x)=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{n+1}}{(n+1)!}=x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+...$ Now, $ x \cdot \ln (1+2x)=x[2x-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^3}{3!}-\dfrac{(2x)^4}{4!}+...]$ Thus, $=2x^2-\dfrac{2^2x^3}{2}+\dfrac{2^3x^{4}}{3}-$ or, $\Sigma_{n=0}^\infty (-1)^n \dfrac{2^{n+1} x^{n+2}}{n+1}$
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