Question

The value of $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n{\sin }^{2k}\left(\frac{r\pi }{2n}\right),$ where $k$ is a non-negative integer, is equal to

• A. $\frac{\left(2k\right)!}{2^{2k}(k!{)}^2}$
• B. $\frac{3k!}{2^{3k}(k!{)}^3}$
• C. $\frac{k!}{2^k(k!{)}^2}$
• D. $\frac{2(k!)}{2^k(k!{)}^2}$

Answer 1

The correct option is A $\frac{\left(2k\right)!}{2^{2k}(k!{)}^2}$

Let $S=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n{\sin }^{2k}\left(\frac{r\pi }{2n}\right)$
$=\underset{0}{\overset{1}{\int }}{\sin }^{2k}\left(\frac{\pi x}{2}\right)dx$
Put $\frac{\pi x}{2}=t\Rightarrow dx=\frac{2}{\pi }dt,$ we have
$S=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2k}\left(t\right)dt$
Using Walli's formula:
$S=\frac{2}{\pi }\frac{(2k-1)(2k-3)(2k-5)\cdots 3\cdot 1}{2k(2k-2)(2k-4)\cdots 4\cdot 2}\cdot \frac{\pi }{2}$
$=\frac{2k(2k-1)(2k-2)(2k-3)\cdots 3\cdot 2\cdot 1}{{\left[2k\right(2k-2\left)\right(2k-4)\cdots 4\cdot 2]}^2}=\frac{\left(2k\right)!}{(2^{k}k!{)}^2}=\frac{\left(2k\right)!}{2^{2k}(k!{)}^2}$