<s>[INST] 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.
 [/INST] If the representation
\begin{align*}
\frac{1}{\sqrt{1-4x}}=\sum_{n\geq 0}\binom{2n}{n}x^n
\end{align*}
for $|x|<\frac{1}{4}$ is supposed to be known, we can also use
\begin{align*}
\sqrt{1-4x}&=\frac{1-4x}{\sqrt{1-4x}}\\
&=\sum_{n \geq 0}\binom{2n}{n}x^n-4\sum_{n\geq 0}\binom{2n}{n}x^{n+1}\\
&=1+\sum_{n\geq 1}\left(\binom{2n}{n}-4\binom{2n-2}{n-1}\right)x^n
\end{align*}
</s>