Answer:
Option B
Explanation:
Let l=$\lim_{n \rightarrow \infty}[\frac{(n+1)(n+2)...3n}{n^{2n}}]^{1/n}$
$\lim_{n \rightarrow \infty}[\frac{(n+1)(n+2)...(n+2n)}{n^{2n}}]^{1/n}$
$\lim_{n \rightarrow \infty}[(\frac{n+1}{n})(\frac{n+2}{n})....(\frac{n+2n}{n})]^{\frac{1}{n}}$
On taking long both sides, we get
$\log I=\lim_{n \rightarrow \infty}\frac{1}{n}[\log {(1+\frac{1}{n})(1+\frac{2}{n}).....(1+\frac{2n}{n})}]$
$\Rightarrow \log l= \lim_{n\rightarrow \infty}\frac{1}{n}$ $[\log (1+\frac{1}{n})+log(1+\frac{2}{n})+.....+log(1+\frac{2n}{n})]$
$\Rightarrow \log l=\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r=1}^{2n}\log(1+\frac{r}{n})$
$\Rightarrow \log l= \int_{0}^{2} \log(1+x)dx$
$\Rightarrow \log l=[\log(1+x).x-\int_{}^{} \frac{1}{1+x}.xdx]_0^2$
$\Rightarrow \log l=[\log(1+x).x]_2^0-\int_{0}^{2} \frac{x+1-x}{1+x}.xdx$
$\Rightarrow \log l=2.\log 3-\int_{0}^{2} (1-\frac{1}{1+X})dx$
$\Rightarrow \log l=2.\log 3-[x-\log\mid1+x\mid]_0^2 $
$Rightarrow \log l=2.\log 3-[ 2-\log3]$
$\Rightarrow \log l=3.\log3-2$
$\Rightarrow \log l=.\log27-2$
$\therefore l=e^{\log 27-2}=27.e^{-2}$
=$\frac{27}{e^{2}}$
