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 - Page A 39: 49

Answer

$2^{n+1}+n^{2}+n-2$

Work Step by Step

Evaluate $\sum\limits_{i =1}^{n}(2i+2^{i})$ $\sum\limits_{i =1}^{n}(2i+2^{i})=\sum\limits_{i =1}^{n}2i+\sum\limits_{i =1}^{n}2^{i}$ Here, $\sum\limits_{i =1}^{n}2^{i}$ shows a geometric series with first term $a=1$ and common ratio $r=2$. Therefore, $\sum\limits_{i =1}^{n}\frac{3}{2^{i-1}}=2[\frac{n(n+1)}{2}]+\frac{2(2^{n}-1)}{2-1}$ $=n(n+1)+\frac{2(2^{n}-1)}{2-1}$ $=n^{2}+n+2^{n+1}-2$ Hence, $\sum\limits_{i =1}^{n}(2i+2^{i})=2^{n+1}+n^{2}+n-2$
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