गुणोत्तर श्रेणी का योगफल निर्दिष्ट पदों तक ज्ञात कीजिए।

मान ज्ञात कीजिए $\sum_{k=1}^{11}\left(2+3^{k}\right)$

$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right) = } \sum\limits_{k = 1}^{11} {\left( 2 \right) + } \sum\limits_{k = 1}^{11} {\left( {{3^k}} \right) = 22 + } \sum\limits_{k = 1}^{11} {{3^k}} $       .........$(1)$

$\sum\limits_{k = 1}^{11} {{3^k} = {3^1} + {3^2} + {3^3} + ........ + {3^{11}}} $

The terms of this sequence $3,3^{2}, 3^{3} \ldots \ldots$ forms a $G.P.$

$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$

$\Rightarrow S_{n}=\frac{3\left[(3)^{11}-1\right]}{3-1}$

$\Rightarrow S_{n}=\frac{3}{2}\left(3^{11}-1\right)$

$\therefore \sum\limits_{k = 1}^{11} {{3^k}}  = \frac{3}{2}\left( {{3^{11}} - 1} \right)$

Substituting this value in equation $(1),$ we obtain

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