$$\mathop {\lim }\limits_{n \to \infty } \frac{{{2^k} + {4^k} + {6^k} + ..... + {{\left( {2n} \right)}^k}}}{{{n^{k + 1}}}},\,k \ne 1,$$         is equal to :

Solution :
$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } {2^k}.\frac{{{1^k} + {2^k} + {3^k} + ..... + {n^k}}}{{{n^{k + 1}}}} \cr & = {2^k}.\mathop {\lim }\limits_{n \to \infty } \sum {\frac{1}{n}} .{\left( {\frac{r}{n}} \right)^k} \cr & = {2^k}.\int_0^1 {{x^k}dx} \cr & = {2^k}.\left[ {\frac{{{x^{k + 1}}}}{{k + 1}}} \right]_0^1 \cr & = \frac{{{2^k}}}{{k + 1}} \cr} $$