Algebra and Trigonometry 10th Edition

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

Chapter 11 - Cumulative Test for Chapters 9-11 - Page 845: 26

Answer

$a_1=\frac{1}{5}$ $a_2=-\frac{1}{7}$ $a_3=\frac{1}{9}$ $a_4=-\frac{1}{11}$ $a_5=\frac{1}{13}$

Work Step by Step

$a_n=\frac{(-1)^{n+1}}{2n+3}$ $a_1=\frac{(-1)^{1+1}}{2(1)+3}=\frac{1}{5}$ $a_2=\frac{(-1)^{2+1}}{2(2)+3}=\frac{-1}{7}=-\frac{1}{7}$ $a_3=\frac{(-1)^{3+1}}{2(3)+3}=\frac{1}{9}$ $a_4=\frac{(-1)^{4+1}}{2(4)+3}=\frac{-1}{11}=-\frac{1}{11}$ $a_5=\frac{(-1)^{5+1}}{2(5)+3}=\frac{1}{13}$
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