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: 12

Answer

$1+3x^2+3x^4+x^6$

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, $(1+x^2)^{3}=1+3x^2+\dfrac{(3)(2)(x^2)^2}{2!}+\dfrac{(3)(2)(1)(x^2)^3}{3!}+0=1+3x^2+3x^4+x^6$
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