Answer
35
Work Step by Step
Reading the graph,
$b_{1}=4$
$b_{2}=2$
$b_{3}=0$
$b_{4}=-2$
$b_{5}=-4$
$\displaystyle \sum_{i=1}^{5}(b_{i}^{2}-1)=$
$=((4)^{2}-1)+((2)^{2}-1)+((0)^{2}-1)+((-2)^{2}-1)+((-4)^{2}-1)$
$=15+3+(-1)+3+15$
$=35$
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.1 - Page 715: 62
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