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,
$\sum\limits_{i =1}^{n}\frac{3}{2^{i-1}}=2[\frac{n(n+1)}{2}]+\frac{(2^{n}-1)}{2-1}$
$=n(n+1)+\frac{(2^{n}-1)}{2-1}$
$=n^{2}+n+2^n-1$
