Browse Questions

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

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

Toolbox:
• $\sum (A+B)=\sum A+\sum B$
• 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)$