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