College Algebra (6th Edition)

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



Work Step by Step

$\sum_{i=1}^{5} (2i^2-3)=(2(1)^2-3)+(2(2)^2-3)+(2(3)^2-3)+(2(4)^2-3)+(2(5)^2-3) =(2-3)+(8-3)+(18-3)+(32-3)+(50-3)=-1+5+15+29+47=95$
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