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: 3

Answer

$1+3x+6x^2+10x^3+.....$

Work Step by Step

Formula to find the binomial series is: $(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$ Here, $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$ Now, $(1-x)^{-3}=1+(-3)(-x)+\dfrac{(-3)(-4)(-x^2)}{2!}+\dfrac{(-3)(-4)(-5)(-x^3)}{3!}+...$ Thus, the first four terms are: $1+3x+6x^2+10x^3+.....$
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