how to prove this by using generating functions coefficient?

Question

First I should calculate the generating function of $(1-x)^{-1/2}$ and by using that prove this identity: $$\displaystyle \sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=4^n.$$ (see this).

Answer 1

We have that (see this page for the details), $$(1+x)^{-1/2} = \sum_{k\geq 0} \binom{-1/2}{k}x^k\quad\mbox{and}\quad (1+x)^{-1}=\sum_{k\geq 0} \binom{-1}{k}x^k.$$ Moreover $\binom{-1}{k}=(-1)^k$ and $$ \binom{-1/2}{k} = (-1)^k \frac{1\cdot 3\cdot 5\cdots (2k-1)}{2^k k!} = \frac{(-1)^k}{4^k}\binom{2k}{k}. $$ Now by Vandermonde's identity, $$\begin{align*} (-1)^n&=[x^{n}](1+x)^{-1}=[x^n]\left((1+x)^{-1/2}\cdot(1+x)^{-1/2}\right)\\ &=\sum_{k=0}^n \frac{(-1)^k}{4^k}\binom{2k}{k}\cdot \frac{(-1)^{n-k}}{4^{n-k}}\binom{2(n-k)}{n-k}\\ &=\frac{(-1)^n}{4^n}\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k} \end{align*}$$ which implies $\displaystyle \sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=4^n.$

Answer 2

Note: I added an explanation of how squaring the series gives the result.

First, by the generalized binomial theorem, $(1+x)^a =\sum_{n=0}^{\infty} \binom{a}{n}x^n $ where $\binom{a}{n} =\dfrac{\prod_{k=0}^{n-1} (a-k)}{n!} $.

If you put $a=-\frac12$, you get

$\begin{array}\\ \binom{-\frac12}{n} &=\dfrac{\prod_{k=0}^{n-1} (-\frac12-k)}{n!}\\ &=\dfrac{\prod_{k=0}^{n-1} (-\frac12)(1+2k)}{n!}\\ &=\dfrac{(-1)^n}{2^n}\dfrac{\prod_{k=0}^{n-1} (2k+1)}{n!}\\ &=\dfrac{(-1)^n}{2^n}\dfrac{\prod_{k=0}^{n-1} (2k+1)(2k+2)}{n!\prod_{k=0}^{n-1} (2k+2)}\\ &=\dfrac{(-1)^n}{2^n}\dfrac{\prod_{k=1}^{2n} k}{n!2^n\prod_{k=0}^{n-1} (k+1)}\\ &=\dfrac{(-1)^n}{2^n}\dfrac{(2n)!}{n!2^nn!}\\ &=\dfrac{(-1)^n(2n)!}{4^nn!^2}\\ &=(-1)^n4^{-n}\binom{2n}{n}\\ \end{array} $

Therefore

$\begin{array}\\ (1-x)^{-1/2} &=\sum_{n=0}^{\infty} \binom{-\frac12}{n}(-x)^n\\ &=\sum_{n=0}^{\infty} 4^{-n}\binom{2n}{n}(-1)^nx^n\\ &=\sum_{n=0}^{\infty} 4^{-n}\binom{2n}{n}x^n\\ \end{array} $

Since $((1-x)^{-1/2})^2 =(1-x)^{-1} =\sum_{n=0}^{\infty} x^n $, by squaring the power series for $((1-x)^{-1/2})^2 $, you get the desired result.

The new stuff:

Here is the squaring:

$\begin{array}\\ ((1-x)^{-1/2})^2 &=\left(\sum_{n=0}^{\infty} 4^{-n}\binom{2n}{n}x^n\right)^2\\ &=\left(\sum_{n=0}^{\infty} 4^{-n}\binom{2n}{n}x^n\right)\left(\sum_{m=0}^{\infty} 4^{-m}\binom{2m}{m}x^m\right)\\ &=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} 4^{-n}\binom{2n}{n}x^n 4^{-m}\binom{2m}{m}x^m\\ &=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} 4^{-n-m}x^{n+m}\binom{2n}{n}\binom{2m}{m}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{n} 4^{-n}x^{n}\binom{2k}{k}\binom{2(n-k)}{n-k}\\ &=\sum_{n=0}^{\infty}x^n4^{-n}\sum_{k=0}^{n} \binom{2k}{k}\binom{2(n-k)}{n-k}\\ \end{array} $

Since $(1-x)^{-1} =\sum_{n=0}^{\infty} x^n $, $1 = 4^{-n}\sum_{k=0}^{n} \binom{2k}{k}\binom{2(n-k)}{n-k} $ or $4^{n}=\sum_{k=0}^{n} \binom{2k}{k}\binom{2(n-k)}{n-k} $.

Note: As in many of my answers, nothing here is original.