CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 103 Marks
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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