University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.7 - Power Series - Exercises - Page 531: 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 have $\Sigma_{n=0}^{\infty} \dfrac{(x-1)^{2n}}{4^n}= \Sigma_{n=0}^{\infty} [\dfrac{x^2-2x+1}{4}]^n=\Sigma_{n=0}^{\infty} (f(x))^n$ The series will converge when $|f(x)|=|\dfrac{x^2-2x+1}{4}| \lt 1$ So, $|\dfrac{(x-1)^2}{4}| \lt 1 \implies -1 \lt x \lt 3$ Now, the sum of a given geometric series is: $S=\dfrac{a}{1-r}=\dfrac{1}{1-\dfrac{x^2-2x+1}{4}}=\dfrac{4}{3+2x-x^2}$ Thus, the Interval of Convergence is: $-1 \lt x \lt 3$ and $\Sigma_{n=0}^{\infty} \dfrac{(x-1)^{2n}}{4^n}=\dfrac{4}{3+2x-x^2}$
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