Finding an expression and proof for $\sqrt{1-4x}$ This is similar to $$\frac{1}{\sqrt{1-4x}}=\sum_{n\geq0}{2n\choose n}x^n$$
However I want to find an expression the same way for $$\sqrt{1-4x}$$ rather than $$\frac{1}{\sqrt{1-4x}}$$
Here's my thoughts so far:
$$(1-4x)^\frac{1}{2}=\sum_{n\geq0}{\frac{1}{2}\choose n}(-4)^nx^n$$
but this is where I get stuck.

Notice that $$\frac{d}{dx}\sqrt{1-4x}=\frac{-2}{\sqrt{1-4x}}$$
So now, $$\frac{-1}{2}\frac{-2}{\sqrt{1-4x}}=\sum_{n\geq 0}\binom{2n}{n}x^n$$ and integrating with respect to $x$ yields $$\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1}=-\frac{1}{2}\sqrt{1-4x} +C $$ so we get that $$\sqrt{1-4x}+C=-2\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1} $$  Letting $x=\frac{1}{4}$, we find that
$$C=-2\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^{n+1}(n+1)}=-\frac{1}{2}\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^n(n+1)}$$
Putting it all together, $$\begin{align}\sqrt{1-4x}&=\frac{1}{2}\sum_{n\geq 0}\binom{2n}{n}\frac{1}{4^n(n+1)} - 2\sum_{n\geq 0}\binom{2n}{n}\frac{x^{n+1}}{n+1}\\&=\frac{1}{2}\sum_{n\geq 0} \binom{2n}{n}\frac{1-(4x)^{n+1}}{4^n(n+1)}\end{align}$$
