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: 34



Work Step by Step

Expanding $(2j+1)^{2}$ and using formulas (3) and (4), we get \begin{equation} \sum_{j=1}^{30}(2j+1)^{2}=\sum_{j=1}^{30}(4j^{2}+4j+1)\end{equation}\begin{equation}=4\sum_{j=1}^{30}j^{2}+4\sum_{j=1}^{30}j+\sum_{j=1}^{30}1\end{equation}\begin{equation}=4\times\frac{30(30+1)(2\times30+1)}{6}+4\times\frac{30(30+1)}{2}+1(30-1+1)=37820+1860+30=39710\end{equation}
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