There are several ways to derive the result; I’ll start with the one using the binomial theorem. What you have isn’t quite right: the $x$ term in $1-x$ is negative, so it should be
$$(1-x)^{-n}=\sum_{k\ge 0}\binom{-n}k(-x)^k=\sum_{k\ge 0}\binom{-n}k(-1)^kx^k\;.$$
For $\binom{-n}k$ you need to know the full definition of the binomial coefficient: for all real $x$ and non-negative integers $k$ we define
$$\binom{x}k=\frac{x^{\underline k}}{k!}=\frac{x(x-1)(x-2)\ldots(x-k+1)}{k!}\;;$$
you can check that this agrees with the more familiar definition when $x$ is a non-negative integer. Now we have
$$\begin{align*}
\binom{-n}k&=\frac{(-n)^{\underline k}}{k!}\\
&=\frac{(-n)(-n-1)(-n-2)\ldots(-n-k+1)}{k!}\\
&=(-1)^k\cdot\frac{n(n+1)(n+2)\ldots(n+k-1)}{k!}\\
&=(-1)^k\cdot\frac{(n+k-1)!}{k!(n-1)!}\\
&=(-1)^k\binom{n+k-1}k\;,
\end{align*}$$
so that
$$(1-x)^{-n}=\sum_{k\ge 0}(-1)^k\binom{n+k-1}k(-1)^kx^k=\sum_{k\ge 0}\binom{n+k-1}kx^k\;,$$
since $(-1)^k(-1)^k=(-1)^{2k}=1$.
Another way is to start from the geometric series
$$\frac1{1-x}=\sum_{k\ge 0}x^k\tag{1}$$
and differentiate repeatedly. If I differentiate $(1-x)^{-1}$ with respect $x$ repeatedly, I get $(1-x)^{-2}$, $2(1-x)^{-3}$, $6(1-x)^{-4}$, $24(1-x)^{-5}$, and in general I have
$$\frac{d^n}{dx^n}(1-x)^{-1}=\frac{n!}{(1-x)^{n+1}}\;;$$
this is easy to prove by induction on $n$.
Differentiating the righthand side of $(1)$ $n$ times with respect to $x$, I get
$$\sum_{k\ge 0}k(k-1)(k-2)\ldots(k-n+1)x^{n-k}\;.$$
Setting $\ell=k-n$, so that $k=\ell+n$, I can rewrite this as
$$\sum_{\ell\ge 0}(\ell+n)(\ell+n-1)\ldots(\ell+1)x^\ell=\sum_{\ell\ge 0}\frac{(\ell+n)!}{\ell!}x^\ell\;.$$
Putting the two pieces together, we see that
$$\frac{n!}{(1-x)^{n+1}}=\sum_{\ell\ge 0}\frac{(\ell+n)!}{\ell!}x^\ell\;,$$
or, after dividing by $n!$,
$$\frac1{(1-x)^{n+1}}=\sum_{\ell\ge 0}\binom{\ell+n}\ell x^\ell\;.$$
Now just replace $n$ by $n-1$ throughout to get
$$(1-x)^{-n}=\sum_{\ell\ge 0}\binom{\ell+n-1}\ell x^\ell\;,$$
as desired.
Best Answer
The idea of $i+j=2k$ is good with the inner summation over $i$: the given sum is $$F(x):=\sum_{k=1}^{\infty}\binom{2k}{k-1}x^{2k}\sum_{i=0}^{2k}\binom{-5/2}{i}\binom{1/2}{2k-i},$$ and the inner sum is $\binom{-2}{2k}=2k+1$ by the Chu-Vandermonde identity. Hence, $$F(x)=\sum_{k=1}^{\infty}(2k+1)\binom{2k}{k-1}x^{2k}=\sum_{k=1}^{\infty}k\binom{2k+1}{k}x^{2k}=x^2 G'(x^2),$$ where $$G(x)=\sum_{k=0}^\infty\binom{2k+1}{k}x^k=\frac{1}{2x}\left(\frac{1}{\sqrt{1-4x}}-1\right)$$ is computed like this, or by recognizing the binomial series, or another way known to you.