Question

If \(\lim_{n\rightarrow \infty}\) \(\frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+....+(nk+n)]=33.\lim_{n\rightarrow \infty}\frac{1}{n^{k+1}}.[1^k+2^k+3^k+....+n^k]\)
then the integral value of k is equal to _______.

Answer 1

Correct Answer: 5

\(\lim_{n\rightarrow \infty}\)\((\frac{n+1}{n})^{k-1} \frac{1}{n}\sum_{r=1}^{n}(k+\frac{r}{n})\) =33

⋅\(\lim_{n\rightarrow \infty}\)\(\frac{1}{n}\sum_{k=1}^{n}(\frac{r}{n})^k\)
\(\Rightarrow \int_{0}^{1}(k+x)dx=33\int_{0}^{1}x^kdx\)
\(\Rightarrow \, \frac{2k+1}{2}=\frac{33}{k+1}\)
\(\Rightarrow K=5\)