Binomial coefficients identity

Question

Prove this identity: $$\binom {N-1/2} {N-1}=\frac{N}{4^{N-1}}\binom{2N-1}{N-1}$$

I was not able to find the proof of this

Answer 1

We obtain \begin{align*} \binom{N-\frac{1}{2}}{N-1}&=\frac{\left(N-\frac{1}{2}\right)\left(N-\frac{3}{2}\right)\cdots3}{(N-1)!}\\ &=\frac{1}{2^{N-1}}\cdot\frac{(2N-1)(2N-3)\cdots 3}{(N-1)!}\\ &=\frac{1}{2^{N-1}}\cdot\frac{(2N-1)!!}{(N-1)!}\tag{1}\\ &=\frac{1}{2^{N-1}}\cdot\frac{(2N)!}{(N-1)!(2N)!!}\tag{2}\\ &=\frac{1}{2^{N-1}}\cdot\frac{(2N)\cdot(2N-1)!}{(N-1)!2^NN!}\tag{3}\\ &=\frac{N}{4^{N-1}}\binom{2N-1}{N-1} \end{align*} and the claim follows.

Comment:

• In (1) we use double factorial notation $(2N-1)!!=(2N-1)\cdot(2N-3)\cdots 5\cdot 3\cdot 1$

• In (2) we use $N!=N!!\cdot(N-1)!!$

• In (3) we use $(2N)!!=2^N\cdot N!$