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: 40



Work Step by Step

Given $$\sum_{j=2}^{30}\left(6 j+\frac{4 j^{2}}{3}\right)$$ 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_{j=2}^{30}\left(6 j+\frac{4 j^{2}}{3}\right)&= 6 \sum_{j=1}^{30} j-6+\frac{4}{3} \sum_{j=1}^{30} j^{2}-\frac{4}{3} ,\ \ \text{Use (1), ( 2)}\\ &= 6\left(\frac{30(30+1)}{2} \right)-6+\frac{4}{3}\left(\frac{30(30+1)(2 (30)+1)}{6}\right)-\frac{4}{3}\\ &=\frac{46168}{3} \end{align*}
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