Question

$\underset{n\to \infty }{\lim }\frac{2^K{+}_4^K+6^K+......+(2n{)}^K}{n^{K+1}}$ is equal to

• A. $2^K$
• B. $\frac{2^K}{K+1}$
• C. $\frac{1}{K+1}$
• D. none of these.

Answer 1

Answer: (A)

Given limit $=\underset{n\to \infty }{\lim }{2}^K\frac{1^K+2^K+3^K+......+n^K}{n^{K+1}}$
$=2^K\underset{n\to \infty }{\lim }\frac{1}{n}\left({\left(\frac{1}{n}\right)}^K+{\left(\frac{2}{n}\right)}^K+{\left(\frac{3}{4}\right)}^K+\dots \dots {\left(\frac{r}{h}\right)}^K+\dots \dots {\left(\frac{n}{n}\right)}^K\right)$
$=2^K{\lim }_{h\to 0}\sum _{rh=h}^{nh}h\cdot {\left(rh\right)}^K$
$=2^K\underset{0}{\overset{1}{\int }}{x}^{K}dx=2^K{\left|\frac{x^{K+1}}{K+1}\right|}_0^1=\frac{2^K}{K+1}$