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: 38


$$\sum_{n=0}^{\infty}\frac{(-1)^n3^nx^n}{4^{n+1}}$$ $|x|\lt4/3$

Work Step by Step

Given $$ f(x)=\frac{1}{4+3x}$$ Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt1\tag{1}$$ By using (1), we get \begin{align*} \frac{1}{4+3x}&=\frac{1}{4(1+(3x/4))}\\ &=\frac{1/4}{(1-(-3x/4))}\\ &=\frac{1}{4} \sum_{n=0}^{\infty}( -3x/4)^{n}\\ &= \frac{1}{4} \sum_{n=0}^{\infty}\frac{(-3)^nx^n}{4^n}\\ &= \sum_{n=0}^{\infty}\frac{(-1)^n3^nx^n}{4^{n+1}}\\ \end{align*} Such that $ |3x/4|\lt 1$, or $|x|\lt4/3$
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