University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 9



Work Step by Step

Since, we know $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+...$ Now, $\dfrac{1}{1+\dfrac{3}{4}x^3}=\dfrac{1}{1-(-\dfrac{3}{4}x^3)}$ or, $=1+(-\dfrac{3}{4}x^3)+(-\dfrac{3}{4}x^3)^2+(-\dfrac{3}{4}x^3)^3+(-\dfrac{3}{4}x^3)^4+...$ or, $=1-\dfrac{3}{4}x^3+\dfrac{9x^6}{16}-\dfrac{27x^9}{64}+\dfrac{81x^{12}}{256}$
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