<s>[INST] Binomial expansion extension to negative powers I know that:
$$\sum_{k=0}^n {n \choose k}a^{n-k}b^k = (a+b)^n$$
But how is this extended to negative powers, for example, I came across the following line of maths, which I struggle to understand:
$$\sum_{r=0}^{\infty}{2r \choose r}x^r = (1-4x)^{-1/2}$$
as well as the following:
$$\sum_{r=0}^{\infty} (2r+1){2r \choose r}x^r = (1-4x)^{-3/2}$$
 [/INST] Both examples are binomial series expansions 
\begin{align*}
(1+x)^{\alpha}=\sum_{r=0}^{\infty}\binom{\alpha}{r}x^r\qquad\qquad  |x|<1  \tag{1}
\end{align*}
convergent for $|x|<1$ with the following definition of binomial coefficients
\begin{align*}
\binom{\alpha}{r}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-r+1)}{r!}\tag{2}
\end{align*}

We obtain from (1)
  \begin{align*}
(1-4x)^{-\frac{1}{2}}&=\sum_{r=0}^{\infty}\binom{-\frac{1}{2}}{r}(-4x)^r\tag{3}\\
&=\sum_{r=0}^{\infty}\frac{(-1)^r}{4^r}\binom{2r}{r}(-4x)^r\\
&=\sum_{r=0}^{\infty}\binom{2r}{r}x^r
\end{align*}
  and similarly
  \begin{align*}
(1-4x)^{-\frac{3}{2}}&=\sum_{r=0}^{\infty}\binom{-\frac{3}{2}}{r}(-4x)^r\tag{4}\\
&=\sum_{r=0}^{\infty}\frac{(-1)^r}{4^r}(2r+1)\binom{2r}{r}(-4x)^r\\
&=\sum_{r=0}^{\infty}(2r+1)\binom{2r}{r}x^r
\end{align*}

Comment:


*

*In (3) we use according to (2)
\begin{align*}
\binom{-\frac{1}{2}}{r}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-3\right)\cdots\left(-\frac{1}{2}-r+1\right)}{r!}\\
&=\frac{(-1)^r}{2^r}\cdot\frac{1\cdot3\cdot5\cdots(2r-1)}{r!}\\
&=\frac{(-1)^r}{2^r}\cdot\frac{(2r-1)!!}{r!}\\
&=\frac{(-1)^r}{2^r}\cdot\frac{(2r)!}{(2r)!!r!}\\
&=\frac{(-1)^r}{4^r}\cdot\frac{(2r)!}{r!r!}\\
&=\frac{(-1)^r}{4^r}\binom{2r}{r}\\
\end{align*}
It is convenient to use double factorials and the formulas $$(2r)!=(2r)!!(2r-1)!!\qquad\text{ and }\qquad(2r)!!=2^rr!$$

*In (4) we find similarly to (3)
\begin{align*}
\binom{-\frac{3}{2}}{r}=\frac{(-1)^r}{4^r}(2r+1)\binom{2r}{r}
\end{align*}
</s>