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 549: 14

Answer

$1-2x+\dfrac{3x^2}{2}-\dfrac{x^3}{2}+\dfrac{x^4}{16}$

Work Step by Step

The formula to determine the binomial series is: $(1+x)^p=1+\Sigma_{k=1}^\infty \dbinom{p}{k}x^k$ Here, $\dbinom{p}{k}=\dfrac{p(p-1)(p-2).....(p-k+1)}{k!}$ Now, $(1-\dfrac{x}{2})^{3}=1+(4)(-\dfrac{x}{2})+\dfrac{(4)(3)(x^2/4)}{2!}+\dfrac{(4)(3)(2)(-x^3/8)}{3!}+\dfrac{(4)(3)(2)(1)(x^4/16)}{4!}=1-2x+\dfrac{3x^2}{2}-\dfrac{x^3}{2}+\dfrac{x^4}{16}$
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