Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.4 The Definition Of Area As A Limit; Sigma Notation - Exercises Set 4.4 - Page 298: 13

Answer

$$2870$$

Work Step by Step

Given $$\sum_{k=1}^{20}(k^2)$$ Since $$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6} $$ Then \begin{align*} \sum_{k=1}^{20}(k^2)&=\frac{(20)(21)(41)}{6} \\ &= 2870 \end{align*}
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