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

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Solution

$\sum_{k = 1}^{11} (2 + 3^{k})$

$= (2 + 3^{1}) + (2 + 3^{2}) + (2 + 3^{3}) + . . . . . + (2 + 3^{11}) .$

$= (2 + 2 + 2 + . . . . 11 t i m e s) + (3 + 3^{2} + 3^{3} + . . . . . 3^{11})$

$= 22 + (3 + 3^{2} + 3^{3} + . . . . + 3^{11}) . . . . (i)$

Now, $3 + 3^{2} + 3^{3} + . . . . . . . . . . . + 3^{11} is a G.P.$

Here, $a = 3 a n d r = \frac{3^{2}}{3} = 3$

$∴ S_{n} = \frac{a [r^{n} - 1]}{r - 1}$
$S_{11} = \frac{3 [3^{11} - 1]}{3 - 1} = \frac{3}{2} (3^{11} - 1)$

Putting the value of $S_{11}$ in $e q . (i),$ we have

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

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