If l i m n → ∞ n + 1 k - 1 n k + 1 ( n k + 1 ) + ( n k + 2 ) + … + n k + n = 33 . lim n…

Given, 

$\underset{n\to \infty }{lim}\frac{{\left(n+1\right)}^{k-1}}{n^{k+1}}\left[(nk+1)+(nk+2)+\dots +\left(nk+n\right)\right]=33$

Taking $\mathrm{LHS}$

$\underset{n\to \infty }{\lim }\frac{{\left(n+1\right)}^{k-1}}{n^{k+1}}\left[nk·n+1+2+\dots +n\right]$

$=\underset{n\to \infty }{\lim }\frac{{\left(n+1\right)}^{{}^{k-1}}}{n^{k+1}}·\left[n^{2}k+\frac{n\left(n+1\right)}{2}\right]$

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

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

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

Now taking $\mathrm{RHS}$

$=33\underset{n\to \infty }{\lim }\frac{1}{n^{k+1}}\left(1^k+2^k+\dots +n^k\right)=33\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{1}{n}{\left(\frac{k}{n}\right)}^k$

$=33{\int }_0^1{x}^{k}dx=\frac{33}{k+1}$

Now given, $\mathrm{LHS}=\mathrm{RHS}$

$\Rightarrow k+\frac{1}{2}=33·\frac{1}{k+1}$

$\Rightarrow \left(2k+1\right)\left(k+1\right)=66$

$\Rightarrow \left(k-5\right)\left(2k+13\right)=0$

$\Rightarrow k=5$ or $-\frac{13}{2}$

So, integral value of $k=5$