Question

The value of the sum $\sum _{k=1}^{\infty }\frac{6^k}{\left(3^k{2}^k\right)\left(3^{k+1}{2}^{k+1}\right)}$ is

• A. 1
• B. 3
• C. 2
• D. 0

Answer 1

The correct option is C 2

Given expression can be written as :
$\frac{3^k}{3^k{2}^k}\frac{3^{k+1}}{3^{k+1}{2}^{k+1}}$
So, required sum
$={lim}_{k\to \infty }[\left(\frac{3^1}{3^1{2}^1}\frac{3^2}{3^2{2}^2}\right)+\left(\frac{3^2}{3^2{2}^2}\frac{3^3}{3^3{2}^3}\right)$
$+...+\left(\frac{3^k}{3^k{2}^k}\frac{3^{k+1}}{3^{k+1}{2}^{k+1}}\right)]$

$={lim}_{k\to \infty }\left\{\frac{3}{32}\frac{3^{k+1}}{3^{k+1}{2}^{k+1}}\right\}$