Question

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

Answer 1

$\sum _{k=1}^{11}(2+3^k)=\sum _{k=1}^{11}{3}^k=2\sum _1^{11}\left(1\right)+\sum _{k=1}^{11}{3}^k=2\left(11\right)+\sum _{k=1}^{11}{3}^k=22+\sum _{k=1}^{11}{3}^k...\left(1\right){\sum }_{k=1}^{11}{3}^k=3^1+3^2+3^3+...3^{11}$
The terms of this sequence $3,3^2,3^{3,}........{.3}^{11}$ forms a G.P.
$\therefore S_n=\frac{a(r^n-1)}{r-1}\Rightarrow S_n=\frac{3[{\left(3\right)}^{11}-1]}{3-1}\Rightarrow S_n=\frac{3}{2}(3^{11}-1)\therefore {\sum }_{k=1}^{11}{3}^k=\frac{3}{2}(3^{11}-1)$
Substituting this value in equation (1), we obtain ${\sum }_{k=1}^{11}{(2+3}^k)=22+\frac{3}{2}(3^{11}-1)$