0

I'm reading a paper, it states that following is true: $$\frac{1}{2}((C_{0} ^{n})^{2}+(C_{1}^{n})^{2}+(C_{3}^{n})^{2}...+(C_{n}^{n})^{2})=\frac{2^{2n}(n-\frac{1}{2})!}{2\sqrt{\pi}n!}\leq \frac{2^{2n}}{2\sqrt{\pi}}$$ where n is positive integer.

My question is: how to achieve $\frac{2^{2n}(n-\frac{1}{2})!}{2\sqrt{\pi}n!}$? I guess it needs helps of following:

  1. $(C_{0}^{n})^{2}+(C_{1}^{n})^{2}+(C_{3}^{n})^{2}...+(C_{n}^{n})^{2})=C_{n}^{2n}$

  2. $\Gamma(\frac{1}{2}+n)=\frac{(2n)!}{4^{n}n!}\sqrt{\pi}=\left( \begin{array}{ll}n-\frac{1}{2}\\n\end{array} \right) n! \sqrt{\pi}$

Could any one tell me about how to do this? thanks

張鶴騰
  • 11

2 Answers2

1

The bracketed sum, that is, $$\sum_{k=0}^{n} \binom {n}{k}^2 =\binom {2n}{n} $$ is called the central binomial coefficient. To see why this is so, see here.

For the proof, see @robjohn's answer here. Hope it helps.

Community
  • 1
0

Thank you for response, I come up with the answer, just some substitutions:

$$\frac{(2n)!}{4^{n}n!}\sqrt{\pi}=\frac{(n-\frac{1}{2})!}{n!(-\frac{1}{2})!} n! \sqrt{\pi}$$ $$\frac{(2n)!}{n!}=\frac{4^{n}(n-\frac{1}{2})!}{\Gamma(\frac{1}{2})}$$ Where $\Gamma(\frac{1}{2})=\sqrt{\pi}$ $$\frac{(2n)!}{n!}=\frac{4^{n}(n-\frac{1}{2})!}{\sqrt{\pi}}$$ Substitute $\frac{(2n)!}{n!}$ back to $\left(\begin{array}{ll} 2n\\n \end{array}\right)=\frac{(2n)!}{n!n!}$: $$\left(\begin{array}{ll} 2n\\n \end{array}\right)=\frac{4^{n}(n-\frac{1}{2})!}{\sqrt{\pi}n!}$$

張鶴騰
  • 11