Evaluate: _ k=1 ^ 11 (2+3^k ) - Vidyadip

Question

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

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Answer

Given: $\sum_{k=1}^{11}\left(2+3^k\right)$
$=\left(2+3^1\right)+\left(2+3^2\right)+\left(2+3^3\right)+\left(2+3^{11}\right)$
$=(2+2+2+\ldots \ldots \ldots 11$ times $)+\left(3+3^2+3^3+\ldots \ldots . .+3^{11}\right)$
$=22+\left(3+3^2+3^3+\ldots \ldots+3^{11}\right) \ldots \ldots \ldots$ (i)
Here $3,3^2, 3^3 \ldots \ldots . ., 3^{11}$ is in G.P.
$\therefore a=3$ and $r=\frac{3^2}{3}=3$
$S _n=\frac{3\left(3^{11}-1\right)}{3-1}=\frac{3}{2}\left(3^{11}-1\right)$
Putting the value of $S _{ n }$ in eq. (i), we get $\sum_{k=1}^{11}\left(2+3^k\right)=22+\frac{3}{2}\left(3^{11}-1\right)$

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