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How would you prove that the generating function of $\binom{2n}{n}$ is $\frac{1}{\sqrt{1-4x}}$?

More precisely, prove that( for $|x|<\frac{1}{4}$ ):

$$\sum^{\infty}_{n=0}x^n\binom{2n}{n}=\frac{1}{\sqrt{1-4x}}$$

Background: I was trying to solve $$S=\sum^{\infty}_{n=0}\frac{(2n+1)!}{8^n(n!)^2}=\sum^{\infty}_{n=0}\frac{(2n+1)}{8^n}\binom{2n}{n}$$ Which if we let $f(x)$ be the generating function in question, would be simply $$f(x)+2xf'(x)$$ With $x=\frac{1}{8}$. Is there a simple proof of the first identity? Wikipedia states it without a proper reference (the reference provided states it without proof). Is there an easier way of calculating $S$? (which is $\sqrt{8}$, by the way)

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  • $\begingroup$ There is a proof by Lagrange inversion at this MSE link. $\endgroup$ Dec 12 '14 at 1:21
  • $\begingroup$ @MarkoRiedel Hm, that's a more general case, but it linked to this. Also, sorry for the wrong previous title, copy-and-paste error. So, close because dupe? $\endgroup$
    – chubakueno
    Dec 12 '14 at 1:27
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Here is a simple derivation of the generating function: start with $$ (1+x)^{-1/2} = \sum_k \binom{-1/2}{k}x^k.$$ Now, \begin{align*} \binom{-1/2}{k} &= \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(\cdots\right)\left(-\frac{1}{2}-k+1\right)}{k!} \\ &= (-1)^k \frac{1\cdot 3\cdot 5\cdots (2k-1)}{2^k k!} \\ &= (-1)^k \frac{(2k)!}{2^k2^kk!k!} \\ &= (-1)^k 2^{-2k}\binom{2k}{k}. \end{align*} Finally replace $x$ by $-4x$ in the binomial expansion, giving $$ (1-4x)^{-1/2} = \sum_k (-1)^k 2^{-2k}\binom{2k}{k}\cdot (-4x)^k = \sum_k \binom{2k}{k}x^k.$$

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Use the binomial theorem. Then $(1-4x)^{-{1 \over 2}} = \sum_{k=0}^\infty \binom{-{1 \over 2}}{k}(-1)^k 4^k x^k$.

$\binom{-{1 \over 2}}{k}(-1)^k 4^k = {\prod_{j=0}^{k-1}((-4)(-{1 \over 2}-j)) \over k!} = {\prod_{j=0}^{k-1}(2(1+2i)) \over k!} = {2^k (2k)!\over 2^kk! k!} = \binom{2k}{k}$

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