Generating Function Proof Prove that $1 \over (1-4x)^2$ generates ${2n \choose n}$,  $n\in N$
I cant find theformula for that type of generating function, i have already tried expanding the combination and trying to expand the generating function. Any Ideas?
 A: First of all, the generating function is not $\frac{1}{(1-4x)^2}$ but is $\frac{1}{\sqrt{1-4x}}$.
Now, using the well known identity $\Gamma(1/2 - n) = \frac{(-4)^n n!}{(2n)!}\sqrt{\pi}$, which can be easily proved via induction by noting $\Gamma(1/2) = \sqrt{\pi}$, we find that
$$\begin{align}
\frac{1}{\sqrt{1-4x}} = (1-4x)^{-1/2} &= \sum_{k=0}^\infty{-1/2 \choose k}(-1)^k (-4)^k x^k
\\&=\sum_{k=0}^\infty \frac{\Gamma(1/2)}{\Gamma(1/2 - k) k!}(-4)^k x^k
\\&=\sum_{k=0}^\infty \frac{\sqrt{\pi}(2k)! }{\sqrt{\pi}(-4)^k (k!)^2}(-4)^k x^k
\\&=\sum_{k=0}^\infty \frac{(2k)!}{(k!)^2} x^k
\\&=\sum_{k=0}^\infty \frac{(2k)!}{(2k-k)!\,k!}x^k
\\&=\sum_{k=0}^\infty {2k \choose k}x^k
\end{align}$$
A: Well, clearly $(1-4x)^{-2}$ is not the generating function, and the reason is this:
$$
(1+x)^{m} = \sum_{n=0}^\infty \binom mn x^n
$$
Where $\binom mn$ now stands for the generalized binomial coefficient, with $m$ being real and $n$ being a positive integer. (At least in our case)
With $m=-2$, what would this look like?
$$
(1+4x)^{-2} = \sum_{n=0}^\infty \binom{-2}{n} (4x)^{n}  
$$
So the coefficients in this case are $\binom{-2}{n} 4^n$, which expands to $$
\frac{-2 \times -3 \times ... \times (-2-n+1)}{n!} \times 4^n
$$
Simplify this for yourself, it is nowhere close to the coefficients you want.
Therefore, your problem as stated is incorrect.
However with $m= -0.5$ we'd get $(1-4x)^{0.5} = \sum_{n=0}^\infty \binom{-0.5}{n}(4x)^{n}$. So, the coefficients come out as $\binom{-0.5}{n} 4^n$. But then:
$$
\frac{-0.5 \times -1.5 \times ... \times (-0.5-n+1)}{n!} \times 4^n = \frac{-1 \times -3 \times ... \times (1-2n)}{n!} \times 2^n \\
= (-1)^{2n}\frac{(2n-1) \times (2n-3) \times ... \times 1}{n!} \times 2^n = \frac{1 \times (2 \times 1) \times 3 \times (2 \times 2) ... \times (2n-1) \times (2 \times n)}{n!n!} = \binom{2n}{n}
$$
and hence the whole thing actually works out.
