expansion of a generating function I found this formula in a book
$$\sqrt{1-4x}=1-2\sum_{n=1}^{\infty}\frac{1}{n} {{2n-2}\choose {n-1}} x^n$$
How can I prove that?

Hint. One may apply the Taylor series expansion to $f(x):=\sqrt{1-4x}$,
$$
f(x)=f(0)+\sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n,\quad |x|<\frac14. \tag1
$$
Then one may prove by induction that
$$
\begin{align}
(\sqrt{u})'=& \frac12\cdot u^{1/2-1}
\\(\sqrt{u})^{(2)}=& \frac12\left(\frac12-1 \right)\cdot u^{1/2-2}
\\(\sqrt{u})^{(3)}=& \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdot u^{1/2-3}
\\\cdots =& \cdots
\\(\sqrt{u})^{(n)}=& \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right)\cdot u^{1/2-n}
\end{align}
$$ that is

$$
\left.(\sqrt{u})^{(n)}\right|_{u=0}= \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right) \tag2
$$ 

From $(2)$ we deduce by the chain rule that, for $n\geq1$,
$$
\begin{align}
\frac{f^{(n)}(0)}{n!}=&\frac1{n!}\left.(\sqrt{1-4x})^{(n)}\right|_{x=0}
\\\\= &\frac{(-4)^n}{n!}\cdot\frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right)
\\\\= &\frac{(-4)^n}{n!}\cdot\frac{1\times(-1)\times(-3)\cdots \times(-(2n-3))}{2^n}
\\\\= &2\cdot\frac{(-1)^n}{n!}\cdot\frac{(-1)^{n-1} 1\times2\times3\cdots \times(2n-3)\times(2n-2)}{1\times2\times3\cdots \times(n-1)}
\\\\= &-\frac2n\cdot\frac{(2n-2)!}{((n-1)!)^2}
\\\\= &-\frac2n\binom{2n-2}{n-1}
\end{align}
$$ which inserted in $(1)$ gives the announced identity.
