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 777: 21

Answer

$a_1=-\frac{1}{2}$ $a_2=\frac{2}{3}$ $a_3=-\frac{3}{4}$ $a_4=\frac{4}{5}$ $a_5=-\frac{5}{6}$

Work Step by Step

$a_n=(-1)^n(\frac{n}{n+1})$ $a_1=(-1)^1(\frac{1}{1+1})=(-1)\frac{1}{2}=-\frac{1}{2}$ $a_2=(-1)^2(\frac{2}{2+1})=1(\frac{2}{3})=\frac{2}{3}$ $a_3=(-1)^3(\frac{3}{3+1})=(-1)\frac{3}{4}=-\frac{3}{4}$ $a_4=(-1)^4(\frac{4}{4+1})=1(\frac{4}{5})=\frac{4}{5}$ $a_5=(-1)^5(\frac{5}{5+1})=(-1)\frac{5}{6}=-\frac{5}{6}$
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