Question

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

Answer 1

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

${\sum }_{k=1}^{11}2=2+2+2+........\left(11times\right)$

$=2\times 11$

$\therefore$ ${\sum }_{k=1}^{11}2=22$

${\sum }_{k=1}^{11}{3}^k=3+3^2+3^3+...{.3}^{11}$

The series is a $G.P$ with first term $a=3,$ common ratio $r=\frac{(3{)}^2}{3}=3$ and number of term $=n$

$S_n=a.\frac{r^n-1}{r-1}$

$S_{11}=3.\frac{3^{11}-1}{3-1}$

$=\frac{3}{2}(3^{11}-1)$

$\therefore$ ${\sum }_{k=1}^{11}{3}^k=\frac{3}{2}(3^{11}-1)$

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