MathematicsMedium210×since 2002Q3283limn→∞((n+1)(n+2)...3nn2n)1n\mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n}}}n→∞lim(n2n(n+1)(n+2)...3n)n1 is equal to:A9e2{9 \over {{e^2}}}e29B3 log 3−23\,\log \,3 - 23log3−2C18e4{{18} \over {{e^4}}}e418D27e2{{27} \over {{e^2}}}e227Check answerSkip