Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 44


For $r$ in $(-2,2)$ The Taylor series for $f(x)=\frac{3x^3}{2-x}$ is $\frac{3x^3}{2}+\frac{3x^4}{4}+\frac{3x^5}{8}+\frac{3x^6}{16}...+\frac{3(x^{n+3})}{2^{n+1}}+...$

Work Step by Step

We are given $f(x)=\frac{3x^3}{2-x}=\frac{\frac{3x^3}{2}}{1-\frac{x}{2}}$ for $r$ in $(-2,2)$ The Taylor series for $\frac{1}{1-x}$ is $1+x+x^2+x^3+...+x^n+...$ The Taylor series for $f(x)=\frac{3x^3}{2-x}$ is $f(x)=\frac{3x^3}{2}.1+\frac{3x^3}{2}(\frac{x}{2})+\frac{3x^3}{2}(\frac{x}{2})^2+\frac{3x^3}{2}(\frac{x}{2})^3...+\frac{3x^3}{2}(\frac{x}{2})^n+...$ $=\frac{3x^3}{2}+\frac{3x^4}{4}+\frac{3x^5}{8}+\frac{3x^6}{16}...+\frac{3(x^{n+3})}{2^{n+1}}+...$
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