`lim_(n tooo) (((n+1)(n+2) . . . 3 n)/(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

B

Explanation

Let `l=underset(n to oo)(lim)(((n+1)*(n+2). . . (3n))/(n^(2n)))^(1/n)`
`=underset(n to oo)(lim)(((n+1)*(n+2). . . (n +2n))/(n^(2n)))^(1/n)`
`=underset(n tooo)(lim)(((n +1)/(n))((n +2)/(n)) . . .((n +2n)/(n)))^(1/n)`
Taking log on both sides , we get
`log l = underset(n tooo)(lim)(1)/(n)[ log{(1+(1)/(n))(1+(2)/(n)) . . .(1+(2n)/(n))}]`
`rArr log l = underset(n tooo)(lim)(1)/(n)`
`[ log(1+(1)/(n))+log(1+(2)/(n))+ . . . + log(1+(2n)/(n))]`
`rArr log l =underset(n tooo)(lim) (1)/(n) sum _(r=1)^(2n)log(1+(r)/(n))`
` rArr log l=int_(0)^(2)log(1+x)dx`
`rArrlog l= [ log (1+x)*x-int(1)/(1+x)*dx]_(0)^(2)`
`rArr logl=[ log (1+x)*x]_(0)^(2)- int_(0)^(2)(x+1-1)/(1+x)dx`
`rArrlog l =2*log 3- int_(0)^(2)(1-(1)/(1+x))dx`
`rArr log l =2* log3-[x-log|1+x|]_(0)^(2)` `rArrlog l =2 *log 3 - [2-log3]`
`rArr log l = 3 * log 3 -2`
`rArr log l = log 27 -2`
`:. l = e^(log27-2)=27.e^(-2)=(27)/(e^(2))`