College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises: 12

Answer

$a_1=\frac{1}{2}$ $a_2=-\frac{2}{3}$ $a_3=\frac{3}{4}$ $a_4=-\frac{4}{5}$ $a_{100}=-\frac{100}{101}$

Work Step by Step

We are given: $a_{n}= (-1)^{n+1}\frac{n}{n+1}= \frac{(-1)^{n+1}n}{n+1}$ We evaluate: $a_1=\frac{(-1)^{2}\ 1}{1+1}=\frac{1}{2}$ $a_2=\frac{(-1)^{3}\ 2}{2+1}=-\frac{2}{3}$ $a_3=\frac{(-1)^{4}\ 3}{3+1}=\frac{3}{4}$ $a_4=\frac{(-1)^{5}\ 4}{4+1}=-\frac{4}{5}$ $a_{100}= \frac{(-1)^{101}\ 100}{101}=-\frac{100}{101}$
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