Answer
$2$
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,
$=(\dfrac{2}{-2})^2+(\dfrac{4}{-4})^2$
$=1+1$
$=2$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
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: 65
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