limn→∞((n+1)(n+2)…3nn2n)1/n\displaystyle\lim_{n\to\infty}\left(\frac{(n+1)(n+2)\dots3n}{n^{2n}}\right)^{1/n}n→∞lim(n2n(n+1)(n+2)…3n)1/n is equal to :
Official answer: B
lnL=lim1n∑k=12nln(1+kn)=∫02ln(1+x) dx=3ln3−2\ln L=\displaystyle\lim\frac1n\sum_{k=1}^{2n}\ln\left(1+\frac kn\right)=\int_0^2\ln(1+x)\,dx=3\ln3-2lnL=limn1k=1∑2nln(1+nk)=∫02ln(1+x)dx=3ln3−2, so L=27e2L=\dfrac{27}{e^2}L=e227.