Question

If $L=\underset{n\to \infty }{lim}(2\times 3^2\times 2^3\times 3^4...\times 2^{n-1}\times 3^n{)}^{\frac{1}{(n^2+1)}}$, then the value of $L^4$ is

• A. 6
• B. 4
• C. 2
• D. 8

Answer 1

The correct option is A 6

$L=\underset{n\to \infty }{lim}{(2\times 3^2\times 2^3\times 3^4\times ...\times 2^{n-1}\times 3^n)}^{\frac{1}{(n^2+1)}}$

$=\underset{n\to \infty }{lim}{(2^{1+3+5+....n-1}\times 3^{2+4+6+...n})}^{\frac{1}{n^2+1}}$

$=\underset{n\to \infty }{lim}{(2^{\frac{n}{4}(2+\frac{(n-1)}{2}2)}\times 3^{\frac{n}{4}(4+\left(\frac{n-1}{2}\right)2)})}^{\frac{1}{n^2+1}}$

$=\underset{n\to \infty }{lim}{(2^{n^2/4}\times 3^{\frac{n^2}{4}+\frac{n}{2}})}^{\frac{1}{n^2+1}}$

$=\underset{n\to \infty }{lim}(2^{\frac{n^2}{4(n^2+1)}}\times 3^{\frac{n^2+2n}{4(n^2+1)}})$

$=\underset{n\to \infty }{lim}(2^{\frac{1}{4(1+\frac{1}{n^2})}}\times 3^{\frac{1+\frac{2}{n}}{4(1+\frac{1}{n^2})}})$

$L=2^{\frac{1}{4(1+0)}}\times 3^{\frac{1+0}{4(1+0)}}=6^{\frac{1}{4}}$

${\left(L\right)}^4=6^{\frac{4}{4}}=6$