Constructing a Hypergeometric Function I am asked to find values for $a,b$ and $c$ such that $$ \frac{1}{2} ((1+x)^{2\alpha}-(1-x)^{2\alpha}) = 2\alpha x\ _2F_1(a,b;c;x^2)$$
I have attempted the following: 
$$\frac{1}{2} ((1+x)^{2\alpha}-(1-x)^{2\alpha}) = \frac{1}{2}\sum^{\infty}_{k=0}\binom{2\alpha}{k}x^k-\sum^{\infty}_{k=0}\binom{2\alpha}{k}(-x)^k \\ = \frac{1}{2} \sum^{\infty}_{k=0}\binom{2\alpha}{k}(x^k-(-x)^k). $$
My lecturer then advised that I should try to rewrite the binomial as Pochammer Symbols which led me to:
$$\frac{1}{2}\sum^{\infty}_{k=0}\frac{(2\alpha -k +1)_k}{k!}(x^k-(-x)^k)$$
We can also see here that for values $k= 2n$ we will recieve a term equal to 0, hence we can rewrite our equation as: 
$$\frac{1}{2}\sum^{\infty}_{n=0}\frac{(2\alpha -(2n+1) +1)_{(2n+1)}}{(2n+1)!}(x^{(2n+1)}-(-x)^{(2n+1)})$$
We simpify to:
$$x\frac{1}{2}\sum^{\infty}_{n=0}\frac{(2\alpha-2n)_{(2n+1)}}{(2n+1)!}(x^{2n}-(-x)^{2n})$$
Now at this point I'm not really sure where to go... I feel like I'm really close and not seeing something or maybe really far away and not aware! 
Either way any help is greatly appreciated!

In the expansion
\begin{align}
f(x)&=\frac{1}{2}\sum_{k=0}^\infty \binom{2\alpha}{k}\left( x^k-(-x)^k \right)\\
&=x\sum_{n=0}^\infty \binom{2\alpha}{2n+1}x^{2n}\\
&=x\sum_{n=0}^\infty c_nX^{n}
\end{align}
with $X=x^2$, the ratio of two successive terms of the series  is
\begin{align}
\frac{c_{n+1}}{c_n}\frac{X^{n+1}}{X^n}&=\frac{\binom{2\alpha}{2n+3}}{\binom{2\alpha}{2n+1}}X\\
&=\frac{(n-\alpha+1/2)(n-\alpha+1)}{(n+3/2)}\frac{X}{n+1}
\end{align}
then, with $a_1=\frac{1}{2}-\alpha$, $ a_2=1-\alpha$ and $ b_1=\frac{3}{2}$ and with $c_0=\binom{2\alpha}{1}=2\alpha$, it gives
\begin{align}
f(x)&=2\alpha x\sum_{n=0}^\infty \frac{(a_1)_n(a_2)_n}{(b_1)_n}\frac{X^n}{n!} \\
&=2\alpha x \ _2F_1\left(\frac{1}{2}-\alpha,1-\alpha;\frac{3}{2};x^2  \right)
\end{align} 
