Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1049: 67

Answer

$80$

Work Step by Step

Here, we have $a_1=-b_1=-4$, $a_2=-b_2=-2$,$a_3=-b_3=0$,$a_4=-b_4=2$,$a_5=-b_5=4$ Thus, $=[(-4)^2+(-2)^2+0^2+2^2+4^2] +[(4)^2+(2)^2+0^2+(-2)^2+(-4)^2]$ $=(16+4+0+4+16)+(16+4+0+4+16)$ $=80$
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