Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 578: 40


$$\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{3 n}}{2^{3 n+4}}$$ $|x|\lt2 $

Work Step by Step

Given $$ f(x)=\frac{1}{16+ 2x^3}$$ Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$ By using (1), we get \begin{align*} \frac{1}{16+ 2x^3} &=\frac{1}{16(1+ x^3/8)}\\ &=\frac{1}{16} \frac{1}{1+ x^3/8} \\ &=\frac{1}{16} \sum_{n=0}^{\infty}(-x^3/8)^{n}\\ &=\frac{1}{16} \sum_{n=0}^{\infty}\frac{(-1) ^nx^ {3n}}{8^n} \\ &= \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{3 n}}{2^{3 n+4}} \end{align*} Such that $ |x/2|\lt 1$, or $|x|\lt2 $
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