## Calculus: Early Transcendentals 8th Edition

$6-3(\frac{1}{2})^{n-1}$
Evaluate $\sum\limits_{i =1}^{n}\frac{3}{2^{i-1}}$ $\sum\limits_{i =1}^{n}\frac{3}{2^{i-1}}=\frac{3}{2^{0}}+\frac{3}{2^{1}}+\frac{3}{2^{2}}+....+\frac{3}{2^{n}}$ $\sum\limits_{i =1}^{n}\frac{3}{2^{i-1}}=\frac{3}{2^{0}}+\sum\limits_{i =1}^{n-1}\frac{3}{2^{i}}$ Here, $\sum\limits_{i =1}^{n-1}\frac{3}{2^{i}}$ shows a geometric series with first term $a=\frac{1}{2}$ and common ratio $r=\frac{1}{2}$. Therefore, $\sum\limits_{i =1}^{n}\frac{3}{2^{i-1}}=3+\frac{\frac{1}{2}(\frac{1}{2}^{n}-1)}{\frac{1}{2}-1}$ $=3+3(1-\frac{1}{2}^{(n-1)})$ $=3+3-3(\frac{1}{2}^{(n-1)})$ Hence, $\sum\limits_{i =1}^{n}\frac{3}{2^{i-1}}=6-3(\frac{1}{2})^{n-1}$