Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.1 Approximating and Computing Area - Exercises - Page 236: 42



Work Step by Step

Given $$\sum_{m=1}^{20}\left(5+\frac{3 m}{2}\right)^{2}$$ Since \begin{align} \sum_{j=1}^{N} j&=1+2+\cdots+N=\frac{N(N+1)}{2} \tag{1}\\ \sum_{j=1}^{N} j^{2}&=1^{2}+2^{2}+\cdots+N^{2}=\frac{N(N+1)(2 N+1)}{6} \tag{2}\\ \sum_{j=1}^{N} j^{3}&=1^{3}+2^{3}+\cdots+N^{3}=\frac{N^{2}(N+1)^{2}}{4} \tag{3} \end{align} Then \begin{align*} \sum_{m=1}^{20}\left(5+\frac{3 m}{2}\right)^{2}&= \sum_{m=1}^{20}\left(25+15 m+\frac{9 m^{2}}{4}\right)\\ &=\sum_{m=1}^{20} 25+\sum_{m=1}^{20} 15 m+\sum_{m=1}^{20} \frac{9 m^{2}}{4}\\ &=25(20)+15 \sum_{m=1}^{20} m+\frac{9}{4} \sum_{m=1}^{20} m^{2}\ \ \text{Use (1), (2)}\\ &=25(20)+15\frac{20(20+1)}{2} +\frac{9}{4} \frac{20(20+1)(2(20)+1)}{6} \\ &=\frac{20215}{2} \end{align*}
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