Answer
$9515$
Work Step by Step
Use the formula $\sum_{k=1}^{n} k^2=\frac{1}{6}n(n+1)(2n+1)$, we have
$\sum_{k=1}^{30} (k^2+2)=\sum_{k=1}^{30} (k^2)+\sum_{k=1}^{30} (2)=\frac{1}{6}(30)(30+1)(2(30)+1)+2(30)=9515$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 11 - Sequences; Induction; the Binomial Theorem - Chapter Review - Review Exercises - Page 859: 14
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