University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 550: 61


$\dfrac{1}{(1+x)^2} =1-2x+3x^2-4x^3+.....; |x| \lt 1$

Work Step by Step

The Taylor series for $\dfrac{-1}{1+x} $ can be defined as: $\dfrac{-1}{1+x}=-\dfrac{-1}{1-(-x)}= -1+x-x^2+x^3-......+x^n+.....; |x| \lt 1$ Consider the given series: $\dfrac{d}{dx}(\dfrac{-1}{1+x}) =\dfrac{d}{dx} [ -1+x-x^2+x^3-......+x^n+.....]$ or, $\dfrac{1}{(1+x)^2} =0+1-2x+3x^2-4x^3+.....$ or, $\dfrac{1}{(1+x)^2} =1-2x+3x^2-4x^3+.....; |x| \lt 1$
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