## Calculus 8th Edition

Published by Cengage

# Appendix E - Sigma Notation - E Exercises - Page A39: 49

#### Answer

$\sum\limits_{i =1}^{n}(2i+2^{i})=2^{n+1}+n^{2}+n-2$

#### 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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