Evaluate $sum_{k=1}^{11}left(2+3^{k} ight)$
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

$\sum_{k=1}^{11}\left(2+3^{k}\right)=22+\frac{3}{2}\left(3^{\prime 1}-1\right)$