` lim_(n to oo) {((n+1)(n+2)(n+3)..3n)/(n^(2n))}^(1/n)` is equal to

Options

A. `(18)/(e^(4))`
B. `(27)/(e^(2))`
C. `(9)/(e^(2))`
D. `3 log 3-2`

Answer

C

Explanation

Let `underset(n to oo) lim{((n+1)(n+2)(n+3)..3n)/(n^(2n))}^(1/n)`
Then,
`logA=underset(n to oo)lim(1)/(n)log{(1+(1)/(n))(1+(2)/(n))(1+(3)/(n))... (1+(2n)/(n))}`
`rArrlogA=underset(n to oo)lim(1)/(n)underset(r=1)overset(2n)sumlog(1+(r)/(n))`
`rArrlogA=underset(n to oo)lim underset(r=1)overset(2n)sumlog(1+(r)/(n))*(1)/(n)`
`rArr A=underset(0)overset(2)intlog (1+x)dx`
`rArr A=underset(0)overset(2)intlog underset(I)((1+x))*underset(II)(1dx)`
`rArr log A=[x log (1+x)]_(0)^(2)-overset(2)underset(0)int(x)/(1+x)dx`
`rArr A=2 log 3(2-log3)=3 log 3-2`
`rArr log A= log9-log e^(2)="log"((9)/(e^(2)))`
`rArr=A=(9)/(e^(2))`