proving complex Binomial Identity Proving the result $\displaystyle \sum^{\infty}_{n=0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$
what i try
$\displaystyle (1+x)^{n}=\sum^{n}_{r=0}\binom{n}{r}x^r$
$\displaystyle (x+1)^n=\sum^{n}_{r=0}\binom{n}{n-r}x^{n-r}$
Campare coefficient of $x^n$ on left and right
$\displaystyle \binom{2n}{n}= \sum^{n}_{k=0}\binom{n}{k}\binom{n}{n-k}$
How do i solve it Help me please 
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
\sum_{n = 0}^{\infty}{2n \choose n}x^{n} & =
\sum_{n = 0}^{\infty}\bracks{{-1/2 \choose n}\pars{-4}^{n}}x^{n} =
\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-4x}^{n}
\\[5mm] & =
\bracks{\vphantom{\Large A}1 + \pars{-4x}}^{\, -1/2} =
\bbx{1 \over \root{1 - 4x}}
\end{align}

Note that
$\ds{{2n \choose n} = {-1/2 \choose n}\pars{-4}^{n}}$.

A: Note that
$$
\begin{align}
\binom{2n}{n}
&=\binom{2n-2}{n-1}\frac{2n(2n-1)}{n^2}\\
&=\binom{2n-2}{n-1}\frac{4n-2}{n}\tag1
\end{align}
$$
Multiply $(1)$ by $4^{-n}$ and set $a_n=4^{-n}\binom{2n}{n}$:
$$
\begin{align}
\overbrace{4^{-n}\binom{2n}{n}}^{\large a_n}
&=4^{-n}\binom{2n-2}{n-1}\frac{4n-2}{n}\\
&=\underbrace{4^{-(n-1)}\binom{2n-2}{n-1}}_{\large a_{n-1}}\frac{n-1/2}{n}\tag2
\end{align}
$$
Therefore, induction and $(2)$ yield
$$
\begin{align}
(-1)^na_n
&=\prod_{k=1}^n\frac{1/2-k}{k}\\
&=\prod_{k=1}^n\frac{-1/2-(k-1)}{k}\\
&=\binom{-1/2}{n}\tag3
\end{align}
$$
Thus,
$$
\binom{2n}{n}=(-4)^n\binom{-1/2}{n}\tag4
$$
Therefore,
$$
\begin{align}
\sum_{n=0}^\infty\binom{2n}{n}x^n
&=\sum_{n=0}^\infty(-4)^n\binom{-1/2}{n}x^n\\[6pt]
&=(1-4x)^{-1/2}\tag5
\end{align}
$$
