_ n ( ( n + 1 ) ( n + 2 ) . ;3n n^ 2n )^ 1 n = - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right) \ldots .\;3n}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}} = $

A
$\frac{9}{{{e^2}}}$
B
$3\log 3 - 2$
C
$\;\frac{{18}}{{{e^4}}}$
✓
$\;\frac{{27}}{{{e^2}}}$

✓

Answer

Correct option: D.

$\;\frac{{27}}{{{e^2}}}$

d
${e^{\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\ell n\left( {1 + \frac{r}{n}} \right)} }} = {e^{\int\limits_0^2 {\ln \left( {1 + x} \right)dx} }}$

$ \Rightarrow {e^{\left( {\left( {x + 1} \right)\left\{ {\ell n\left( {x + 1} \right) - 1} \right\}} \right)_0^2}} = {e^{3\ell n3 - 2}} = \frac{{27}}{{{e^2}}}$

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