# 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

• Mar 20, 2019 at 3:58

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \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 $$\ds{{2n \choose n} = {-1/2 \choose n}\pars{-4}^{n}}$$.
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}