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Q.

The value of the following limit

${(\frac{(n + 1) (n + 2) (n + 3) \dots \dots . 3 n}{n^{2 n}})}^{\frac{1}{n}}$ is

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a

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

b

27 e^{2}

c

- \frac{27}{e^{2}}

d

- 27 e^{2}

answer is A.

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Detailed Solution

Let y

=

{(\frac{(n + 1) (n + 2) (n + 3) \dots \dots . 3 n}{n^{2 n}})}^{\frac{1}{n}}

Taking log on both sides, we get
ln y =

\frac{1}{n} (\frac{(n + 1) (n + 2) (n + 3) \dots \dots . (n + 2 n)}{n . n . n \dots \dots . . 2 n times})

ln y =

\frac{1}{n} (\frac{(n + 1)}{n} \frac{(n + 2)}{n} \frac{(n + 3)}{n} \dots \dots \frac{(n + 2 n)}{n})

ln y =

\frac{1}{n} (\frac{(n + 1)}{n} + \frac{(n + 2)}{n} + \frac{(n + 3)}{n} \dots \dots + \frac{(n + 2 n)}{n})

ln y =

\frac{1}{n}

\sum_{r = 1}^{2 n} (\frac{n + r}{n})

ln y =

\frac{1}{n}

\sum_{r = 1}^{2 n} (1 + \frac{r}{n})

Put

\frac{r}{n} = x

, then

\frac{1}{n} dr = dx

Also, when r = 1, then

x = \frac{1}{n} = 0

r = 2 n,

then

x = 2

n \to \infty,

then

x \to 0

Thus, ln y =

\frac{1}{n}

\sum_{r = 1}^{2 n} (1 + \frac{r}{n})

= \int_{0}^{2} \ln (1 + x) dx

We know that,

\int_{}^{} \ln ln x dx = x \ln \ln (x) - x

⇒

\int_{0}^{2} \ln (1 + x) dx = {[(1 + x) \ln \ln (1 + x) - (1 + x)]}_{0}^{2}

= 3 \ln \ln 3 - 3 - 0 \ln \ln 1 + (1 + 0)

= 3 \ln \ln 3 - 2

= \ln \ln 3^{3} - \ln \ln e^{2}

= \ln \ln \frac{27}{e^{2}}

Thus,

\ln ln y = \ln \ln \frac{27}{e^{2}}

⇒

y = \frac{27}{e^{2}}

Hence, the correct option is 1.

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