If $\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{(\mathrm{n}+1{)}^{\mathrm{k}-1}}{{\mathrm{n}}^{\mathrm{k}+1}}\left[\right(\mathrm{nk}+1)+(\mathrm{nk}+2)+\dots +(\mathrm{nk}+\mathrm{n}\left)\right]=33$. $\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{1}{{\mathrm{n}}^{\mathrm{k}+1}}\cdot \left[1^{\mathrm{k}}+2^{\mathrm{k}}+3^{\mathrm{k}}+\dots +{\mathrm{n}}^{\mathrm{k}}\right]$, then the integral value of k is equal to _______.

LHS

$\begin{array}{l}\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{(\mathrm{n}+1{)}^{\mathrm{k}-1}}{{\mathrm{n}}^{\mathrm{k}+1}}[\mathrm{nk}\cdot \mathrm{n}+1+2+\dots +\mathrm{n}]\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{(\mathrm{n}+1{)}^{\mathrm{k}-1}}{{\mathrm{n}}^{\mathrm{k}+1}}\cdot \left[{\mathrm{n}}^2\mathrm{k}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]\\ \left.=\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{(\mathrm{n}+1{)}^{\mathrm{k}-1}\cdot {\mathrm{n}}^2\left(\mathrm{k}+\frac{\left(1+\frac{1}{\mathrm{n}}\right)}{2}\right)}{{\mathrm{n}}^{\mathrm{k}+1}}\right)\\ \Rightarrow \underset{\mathrm{n}\to \mathrm{\infty }}{lim} \left(1+\frac{1}{\mathrm{n}}\right)\left(\mathrm{k}+\frac{\left(1+\frac{1}{\mathrm{n}}\right)}{2}\right)\end{array}$

$\Rightarrow \left(\mathrm{k}+\frac{1}{2}\right)$

RHS

$\Rightarrow \underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{1}{{\mathrm{n}}^{\mathrm{k}+1}}\left(1^{\mathrm{k}}+2^{\mathrm{k}}+\dots +{\mathrm{n}}^{\mathrm{k}}\right)=\frac{1}{\mathrm{k}+1}$

LHS = RHS

$\begin{array}{l}\Rightarrow \mathrm{k}+\frac{1}{2}=33\cdot \frac{1}{\mathrm{k}+1}\\ \Rightarrow (2\mathrm{k}+1)(\mathrm{k}+1)=66\\ \Rightarrow (\mathrm{k}-5)(2\mathrm{k}+13)=0\\ \Rightarrow \mathrm{k}=5\text{ or }-\frac{13}{2}\end{array}$