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: 39



Work Step by Step

\begin{equation} \sum_{j=0}^{50}j(j-1)=\sum_{j=0}^{50}(j^{2}-j)\end{equation}\begin{equation}=(0^{2}-0)+\sum_{j=1}^{50}(j^{2}-j)\end{equation}\begin{equation}=\sum_{j=1}^{50}j^{2}-\sum_{j=1}^{50}j \end{equation}\begin{equation}=\frac{50(50+1)(2\times50+1)}{6}-\frac{50(50+1)}{2}=41,650\end{equation}
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