How can I prove that $\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^{\infty} \binom {2n} {n} x^n$?
I know the I should use Newton's generalized binomial theorem which says that,
$$(1+z)^a=\sum_{n=0}^{\infty} \binom {a} {n}z^n=1+\binom{a}{1}z+\binom{a}{2}z^2+...$$
But how can I use it in this special case?
On
It is possible to use the catalan number generating function $C(z)=\sum_{n\ge 0}c_nz^{2n}$ to derive the generating function $f(z)$ for NE lattice paths from $(0,0)$ to $(n,n)$.
The catalan number generating function is:
$$C(z)=\frac{1}{2z^2}\left(1-\sqrt{1-4z^2}\right)\tag{1}$$
which gives, as it's $z^{2n}$ coefficient, the number of NE lattice paths starting at $(0,0)$, that do not exceed the diagonal $y=x$, and terminate at $(n,n)$.
The total number of NE lattice paths starting at $(0,0)$ and that terminate at a diagonal coordinate $(n,n)$ is found by splitting NE lattice paths into $k=0,1,2,3,\ldots$ catalan paths joined by $2$ NE steps which intersect the diagonal (plus one step at the start and at the end). Since each catalan path may be either above or below the diagonal there is a multiplication factor of 2 for each catalan path generated by $C(z)$.
The generating function $f(z^2)=\sum_{n\ge 0}\binom{2n}{n}z^{2z}$ for all paths from $(0,0)$ to $(n,n)$ is therefore:
$$\begin{align}f(z^2)=(2C(z)z^2)^0+(2C(z)z^2)^1+&(2C(z)z^2)^2+(2C(z)z^2)^3+\cdots +(2C(z)z^2)^k+\cdots \\&=\frac{1}{1-2C(z)z^2}\tag{2}\end{align}$$
inputting $(1)$ into the right hand side of $(2)$ gives:
$$f(z^2)=\frac{1}{\sqrt{1-4z^2}}$$ $$\implies f(z)=\frac{1}{\sqrt{1-4z}}\tag*{$\blacksquare$}$$
\begin{eqnarray*} \binom{-1/2}{n} (-4)^n = \frac{(-1/2)(-3/2) \cdots (-1/2-(n-1))}{n!} (-4)^n=\frac{(-1)^n(2n-1)!!}{2^n n!} \frac{(2n)!!}{2^n n!} (-4)^n = \binom{2n}{n}. \end{eqnarray*}