Answer
$133$
Work Step by Step
$a_1=1^2+7=8$
$a_2=2^2+7=11$
$a_3=3^2+7=16$
$a_4=4^2+7=23$
$a_5=5^2+7=32$
$a_6=6^2+7=43$
Hence $\sum_{n=1}^6(n^2+7)=8+11+16+23+32+43=133$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 12 Sequences and Series - Chapter Review - Page 840: 5
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