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: 64



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, $=[-4+3(4)]+[-2+3(2)]+[0+3(0)]+[2+3(-2)]+[4+3(-4)]$ $=8+4+0-4-8$ $=0$
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