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

$\begin{array}{1 1} 2+\large\frac{3}{2}(3^{10}-1) \\ 2+\large\frac{3}{2}(3^{11}-1) \\22+\large\frac{3}{2}(3^{11}-1) \\ 22+\large\frac{3}{2}(3^{10}-1)\end{array} $

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  • $\sum (A+B)=\sum A+\sum B$
  • Sum of $n$ terms of a G.P.$=a.\large\frac{r^n-1}{r-1}$
$\sum\limits_{k=1}^{11} (2+3^k)$=$\sum\limits_{k=1}^{11} 2+\sum\limits_{k=1}^{11} 3^k$
$\sum\limits_{k=1}^{11} 2=2+2+2+...........(11\:times)$
$=2\times 11=22$
$\sum\limits_{k=1}^{11} 3^k=3+3^2+3^3+........3^{11}$
This series is a G.P. with first term= $a=3,$ common ratio $=r=\large\frac{3^2}{3}$$=3$
and number of terms $=n$
We know that sum of $n$ terms of a G.P$=a.\large\frac{r^n-1}{r-1}$
$\therefore\:\sum\limits_{k=1}^{11} 3^k=S_{11}=3.\large\frac{3^{11}-1}{3-1}=\frac{3}{2}$$(3^{11}-1)$
$\Rightarrow\:\sum\limits_{k=1}^{11} (2+3^k)=22+\large\frac{3}{2}$$(3^{11}-1)$
answered Feb 26, 2014 by rvidyagovindarajan_1

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