Answer
$6-3(\frac{1}{2})^{n-1}$
Work Step by Step
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, $\sum\limits_{i =1}^{n}\frac{3}{2^{i-1}}=6-3(\frac{1}{2})^{n-1}$
