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: 41



Work Step by Step

Given $$\sum_{m=1}^{30}(4-m)^{3}$$ 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}^{30}(4-m)^{3}&= \sum_{m=1}^{30}\left(64-m^{3}-48 m+12 m^{2}\right)\\ &=\sum_{m=1}^{30} 64-\sum_{m=1}^{30} m^{3}-\sum_{m=1}^{30} 48 m+\sum_{m=1}^{30} 12 m^{2}\\ &=\sum_{m=1}^{30} 64-\sum_{m=1}^{30} m^{3}-48 \sum_{m=1}^{30} m+12 \sum_{m=1}^{30} m^{2}, \ \text{Use (1), (2), (3)}\\ &= (64)(30)-\frac{30^{2}(30+1)^{2}}{4} -48\frac{30(30+1)}{2}+12 \frac{30(30+1)(2 (30)+1)}{6} \\ &= -123165 \end{align*}
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