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 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 22

Answer

$\Sigma_{n=0}^\infty (n+1)(n+2)x^{(n)}$

Work Step by Step

Consider the Taylor Series for $\dfrac{1}{(1-x)^2}$ as follows: $\dfrac{1}{(1-x)^2}=1+2x+3x^2+.....nx^{n-1}$ Then, we have $ \dfrac{d}{dx}[\dfrac{1}{(1-x)^2}]=\dfrac{d}{dx}[1+2x+3x^2+.....nx^{n-1}]$ or, $=2+6x+12x^2+....n(n+1)x^{n-1}$ This implies that $\dfrac{2}{(1-x)^3}=\Sigma_{n=0}^\infty (n+1)(n+2)x^{(n)}$
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