Question

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

• A. 5.00
• B. 05
• C. 5.0
• D. 5

Answer 1

Answer: (A), (B), (C), (D)

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

$\Rightarrow {\int }_0^1(k+x)dx=33{\int }_0^1{x}^{k}dx$

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

$\Rightarrow k=5$