Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX E - Sigma Notation - E Exercises: 50

Answer

$\sum\limits_{i =1}^{m}[\sum\limits_{j =1}^{n}(i+j)]=\frac{1}{2}mn(m+n+2)$

Work Step by Step

Evaluate $\sum\limits_{i =1}^{m}[\sum\limits_{j =1}^{n}(i+j)]$ $\sum\limits_{i =1}^{m}[\sum\limits_{j =1}^{n}(i+j)]=i\sum\limits_{j =1}^{n}1+\sum\limits_{j =1}^{n}j$ $=n[\frac{m(m+1)}{2}]+m[\frac{n(n+1)}{2}]$ $=\frac{1}{2}mn[(m+1)+(n+1)]$ Hence, $\sum\limits_{i =1}^{m}[\sum\limits_{j =1}^{n}(i+j)]=\frac{1}{2}mn(m+n+2)$
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