Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - Review Exercises - Page 312: 27

Answer

$\Sigma^{20}_{i=1} (i+1)^{2}$ =3310

Work Step by Step

$\Sigma^{20}_{i=1} (i+1)^{2}$ = $\Sigma^{20}_{i=1} (i^{2}+2i+1)$ =$\Sigma^{20}_{i=1}i^{2} + \Sigma^{20}_{i=1} 2i +\Sigma^{20}_{i=1} 1$ Using the properties of summation $\Sigma^{n}_{i=1} i=\frac{n(n+1)}{2}$ and $\Sigma^{n}_{i=1} i^{2}=\frac{n(n+1)(2n+1)}{6}$, $\Sigma^{20}_{i=1}i^{2} + \Sigma^{20}_{i=1} 2i +\Sigma^{20}_{i=1} 1$=$\frac{20(21)(41)}{6}+2\frac{20(21)}{2} +20$ =3310
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