Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 42


$\frac{1}{\sqrt 2}(x^{2}-\frac{1}{4}x^{3}+\frac{3}{32}x^{4}-....)$

Work Step by Step

$\frac {x^{2}}{\sqrt{(2+x)}}=\frac{x^{2}}{\sqrt 2}\frac{1}{\sqrt {1+\frac{x}{2}}}$ $=\frac{x^{2}}{\sqrt 2}(1+\frac{x}{2})^{-1/2}$ $=\frac{1}{\sqrt 2}( x^{2}+(-\frac{1}{2})(\frac{x^{3}}{2})+\frac{(\frac{-1}{2})(\frac{-3}{2})}{2^{2}2!}x^{4}+\frac{(\frac{-1}{2})(\frac{-3}{2})(\frac{-5}{2})}{2^{3}3!}x^{5}+....)$ $=\frac{1}{\sqrt 2}(x^{2}-\frac{1}{4}x^{3}+\frac{3}{32}x^{4}-....)$
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