Answer
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Work Step by Step
Given $a_j=j^2+1$, we have:
$a_1=1^2+1=2$
$a_2=2^2+1=5$
$a_3=3^2+1=10$
$a_4=4^2+1=17$
$a_5=5^2+1=26$
$a_6=6^2+1=37$
The sum is
$\sum_1^6 a_j=2+5+10+17+26+37=97$
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 9 - Section 9.6 - Conic Sections in Polar Coordinates - Exercise Set - Page 1032: 61
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