Answer
0
Work Step by Step
We evaluate the sum:
$\sum_{j=1}^{100}(-1)^{j}=(-1)^{1}+(-1)^{2}+(-1)^{3}+(-1)^{4}+(-1)^{5}+...+(-1)^{99}+(-1)^{100}=-1+1-1+1-1+...-1+1=0$
(Since we have an even number of terms, the $+1$ and $-1$ cancel out.)
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
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 - Page 601: 50
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