Question:
Evaluate $\sum_{k=1}^{11}\left(2+3^{k}\right)$
Solution:
$\sum_{k=1}^{11}\left(2+3^{k}\right)=\sum_{k=1}^{11}(2)+\sum_{k=1}^{11} 3^{k}=2(11)+\sum_{k=1}^{11} 3^{k}=22+\sum_{k=1}^{11} 3^{k}$ $\ldots(1)$
$\sum_{k=1}^{11} 3^{k}=3^{1}+3^{2}+3^{3}+\ldots+3^{11}$
The terms of this sequence $3,3^{2}, 3^{3}, \ldots$ forms a G.P.
$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$
$\Rightarrow \mathrm{S}_{\|}=\frac{3\left[(3)^{\|}-1\right]}{3-1}$
$\Rightarrow S_{11}=\frac{3}{2}\left(3^{11}-1\right)$
$\therefore \sum_{k=1}^{11} 3^{k}=\frac{3}{2}\left(3^{11}-1\right)$
Substituting this value in equation (1), we obtain
$\sum_{k=1}^{11}\left(2+3^{k}\right)=22+\frac{3}{2}\left(3^{\prime 1}-1\right)$