Answer
$95$
Work Step by Step
$\sum_{i=1}^{5} (2i^2-3)=(2(1)^2-3)+(2(2)^2-3)+(2(3)^2-3)+(2(4)^2-3)+(2(5)^2-3) =(2-3)+(8-3)+(18-3)+(32-3)+(50-3)=-1+5+15+29+47=95$
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 - Summary, Review, and Test - Review Exercises - Page 788: 8
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