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 235: 24


$$\sum_{i=2}^{5} (i^2+i)$$

Work Step by Step

Given $$\left(2^{2}+2\right)+\left(3^{2}+3\right)+\left(4^{2}+4\right)+\left(5^{2}+5\right)$$ The first term is $ \left(2^{2}+2\right)$, the last term is $\left(5^{2}+5\right)$, and we observe that the power is fixed at $2$; thus $$\left(2^{2}+2\right)+\left(3^{2}+3\right)+\left(4^{2}+4\right)+\left(5^{2}+5\right)=\sum_{i=2}^{5} (i^2+i)$$
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