Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.1 - Sequences and Series - 11.1 Exercises - Page 778: 62


$a_0=-1$ $a_1=-\frac{1}{6}$ $a_2=-\frac{1}{120}$ $a_3=-\frac{1}{5040}$ $a_4=-\frac{1}{362880}$

Work Step by Step

$a_n=\frac{(-1)^{2n+1}}{(2n+1)!}$ $a_0=\frac{(-1)^{2(0)+1}}{[2(0)+1]!}=\frac{(-1)^1}{1!}=-1$ $a_1=\frac{(-1)^{2(1)+1}}{[2(1)+1]!}=\frac{(-1)^3}{3!}=-\frac{1}{6}$ $a_2=\frac{(-1)^{2(2)+1}}{[2(2)+1]!}=\frac{(-1)^5}{5!}=-\frac{1}{120}$ $a_3=\frac{(-1)^{2(3)+1}}{[2(3)+1]!}=\frac{(-1)^7}{7!}=-\frac{1}{5040}$ $a_4=\frac{(-1)^{2(4)+1}}{[2(4)+1]!}=\frac{(-1)^9}{9!}=-\frac{1}{362880}$
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