Question

The value of $101\underset{n\to \infty }{lim}\left(\frac{1+2^{100}+...+n^{100}}{n^{101}}\right)$ is

• A. 2
• B. 1
• C. 6
• D. 5

Answer 1

The correct option is B 1

$101.\underset{n\to \infty }{lim}\left(\frac{1+2^{100}+...+n^{100}}{n^{101}}\right)=101$

We know that, $\underset{n\to \infty }{lim}\sum _{r=1}^{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f\left(x\right)dx$

${\int }_0^1{x}^{100}dx=101.{\left(\frac{x^{101}}{101}\right)}_0^1=1$