Finding Binomial expansion of a radical I am having trouble finding the correct binomial expansion for $\dfrac{1}{\sqrt{1-4x}}$:
Simplifying the radical I get: $(1-4x)^{-\frac{1}{2}}$
Now I want to find ${n\choose k} = {\frac{-1}{2}\choose k}$
\begin{align}
{\frac{-1}{2}\choose k} &= \dfrac{\frac{-1}{2}(\frac{-1}{2}-1)(\frac{-1}{2}-2)\ldots(\frac{-1}{2}-k+1)}{k!} \\
&= (-1)^k\dfrac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)\ldots(\frac{1}{2}+k-1)}{k!} \\
&=(-1)^k{\frac{1}{2}+k-1\choose k}
\end{align}
Now, applying Newton's Generalized Binomial Theorem I get:
\begin{equation}\sum\limits_{k=0}^\infty(-1)^k{\frac{1}{2}+k-1\choose k}(4x)^k
\end{equation}
Is this correct? it seems odd to me since the top number of the combination is always less than the bottom number.
Furthermore, I would like to write $\left(\dfrac{1}{\sqrt{1-4x}}\right)^2$ as an infinite sum like: 
\begin{equation}\sum\limits_{k=0}^\infty\sum\limits_{i=0}^k(expression)x^k
\end{equation}
 I should be able to square the entire sum for what I derived above, and get what I know from calculus: 
\begin{equation}
 \left(\dfrac{1}{\sqrt{1-4x}}\right)^2 = \dfrac{1}{1-4x} = \sum\limits_{k=0}^\infty(4x)^k
\end{equation}
The trouble is finding the expression for the double sum.
All help is greatly appreciated!
EDIT: With the help from below we should get the double sum to be: $\sum\limits_{k=0}^\infty\sum\limits_{i=0}^k{k\choose i}^2x^k$

Also, generalizing what Andre Nicolas wrote,
\begin{align}
{\frac{-1}{2}\choose k} &= \dfrac{\frac{-1}{2}(\frac{-1}{2}-1)(\frac{-1}{2}-2)\ldots(\frac{-1}{2}-k+1)}{k!} \\
&= (-1)^k\dfrac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)\ldots(\frac{1}{2}+k-1)}{k!} \\
&=(-1)^k \frac{1\  3\ 5 ... (2k-1)}{2^k k!} \\
&=(-1)^k \frac{1\ 2\  3\ 4\ 5 ... (2k-1)(2k)}{2^k k!(2\ 4\ ...(2k))} \\
&=(-1)^k \frac{(2k)!}{2^k k!2^k(1\ 2\ ...k)} \\
&=(-1)^k \frac{(2k)!}{4^k k!^2} \\
&= (-1)^k {2k \choose k} \big/4^{k} \\
\end{align}
This is a standard bit of manipulation
that, combined with Stirling's factorial approximation,
allows good approximations for this expression
to be derived.
