Question

The value of $\displaystyle\sum_{k=1}^{11}\left(2+3^k\right)$ is

• A. $\frac{3\left(3^{11}-1\right)}{2}+33$
• B. $\frac{3\left(3^{10}+1\right)}{2}+22$
• C. $\frac{3\left(3^{11}+1\right)}{2}+33$
• D. $\frac{3\left(3^{11}-1\right)}{2}+22$

Answer 1

Answer: (D)

$\displaystyle \sum_{k=1}^{11} 2+\sum_{k=1}^{11} 3^k=2 \times 11+\left(3^1+3^2+3^3+\cdots+3^{11}\right) $
$=22+\frac{3\left(3^{11}-1\right)}{3-1}=22+\frac{3\left(3^{11}-1\right)}{2}$
$ {\left[\because S_n=\frac{a\left(r^n-1\right)}{r-1} \text { as } r>1\right] }$