Answer
$1560$
Work Step by Step
I know that $\sum_{k=1}^{n} (k^2)=\frac{n(n+1)(2n+1)}{6}.$
Hence $\sum_{k=1}^{16} (k^2+4)=\sum_{k=1}^{16} (k^2)+\sum_{k=1}^{16} (4)=\frac{16(16+1)(32+1)}{6}+16(4)=1496+63=1560$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.1 Sequences - 12.1 Assess Your Understanding - Page 808: 75
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