Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.7 - Power Series - Exercises 10.7 - Page 613: 43

Answer

Interval of Convergence: $-1 \lt x \lt 3$ and $\Sigma_{n=0}^{\infty} \dfrac{(x-1)^{2n}}{4^n}=\dfrac{4}{3+2x-x^2}$

Work Step by Step

We notice that $\Sigma_{n=0}^{\infty} \dfrac{(x-1)^{2n}}{4^n}= \Sigma_{n=0}^{\infty} [\dfrac{x^2-2x+1}{4}]^n=\Sigma_{n=0}^{\infty} (f(x))^n$ Remember that the series will converge when $|f(x)|=|\dfrac{x^2-2x+1}{4}| \lt 1$ This implies that $|\dfrac{(x-1)^2}{4}| \lt 1 $ and $ -1 \lt x \lt 3$ This means that the interval of convergence is: $ \ln 3 \lt x \lt \ln 5$ Next, we will find the sum of the given geometric series $S=\dfrac{a}{1-r}=\dfrac{1}{1-\dfrac{x^2-2x+1}{4}}=\dfrac{4}{3+2x-x^2}$ $\implies \Sigma_{n=0}^{\infty} \dfrac{(x-1)^{2n}}{4^n}=\dfrac{4}{3+2x-x^2}$
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