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: 13



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, $(1-2x)^{3}=1+3(-2x)+\dfrac{(3)(2)(4x^2)}{2!}+\dfrac{(3)(2)(1)(-8)x^3}{3!}=1-6x+12x^2-8x^3$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.