Bounding a sum of products of binomial coefficients I am trying to understand the following sums for $k\le n$ :
$$
\sum_{s=0}^{k} \begin{pmatrix} 2n-s/2\\ s\end{pmatrix}\begin{pmatrix} 2n-3s/2\\ k-s\end{pmatrix}
$$
$$
\sum_{s=0}^{k} \begin{pmatrix} 2n-s\\ s\end{pmatrix}\begin{pmatrix} 2n-s\\ k-s\end{pmatrix}
$$
More precisely, I want to know if there is an $\alpha$, respectively $\beta$, such that for any $\epsilon > 0$ and sufficiently big $n$, we have
$$
\begin{pmatrix} 4n-(\alpha+\epsilon) k\\ k\end{pmatrix} 
\le \sum_{s=0}^{k} \begin{pmatrix} 2n-s/2\\ s\end{pmatrix}\begin{pmatrix} 2n-3s/2\\ k-s\end{pmatrix}
\le \begin{pmatrix} 4n-(\alpha-\epsilon) k\\ k\end{pmatrix},
$$
respectively
$$
\begin{pmatrix} 4n-(\beta+\epsilon) k\\ k\end{pmatrix} 
\le \sum_{s=0}^{k} \begin{pmatrix} 2n-s\\ s\end{pmatrix}\begin{pmatrix} 2n-s\\ k-s\end{pmatrix}
\le \begin{pmatrix} 4n-(\beta-\epsilon) k\\ k\end{pmatrix} .
$$
I am also fine with having rational function factors or rational powers of such factors in the bounding terms. 
One can see 
$$
\begin{pmatrix} 4n-2k\\ k\end{pmatrix} \le \sum_{s=0}^{k} \begin{pmatrix} 2n-k/2\\ s\end{pmatrix}\begin{pmatrix} 2n-3k/2\\ k-s\end{pmatrix}
\le \sum_{s=0}^{k} \begin{pmatrix} 2n-s/2\\ s\end{pmatrix}\begin{pmatrix} 2n-3s/2\\ k-s\end{pmatrix}\le \sum_{s=0}^{k} \begin{pmatrix} 2n\\ s\end{pmatrix}\begin{pmatrix} 2n\\ k-s\end{pmatrix}
= \begin{pmatrix} 4n\\ k\end{pmatrix},
$$
hence if there is such $\alpha$ then $0\le \alpha \le 2$.

Suppose that $k$ is fixed. The sums represent polynomials of degree $k$ in $n$. To answer your question, it's enough to compute two first leading terms of these polynomials. Let me do that for the second sum (the first sum is treated similarly), replacing $2n$ with $n$.
We have
\begin{split}
S(n,k)&:=\sum_{s=0}^{k} \binom{n-s}s \binom{n-s}{k-s} \\
&= \sum_{s=0}^{k} \frac{n^s - \frac{s(3s-1)}2n^{s-1}}{s!}\cdot \frac{n^{k-s} - \frac{(k-s)(k+s-1)}2n^{k-s-1}}{(k-s)!} + O(n^{k-2})\\
&=\frac1{k!}\sum_{s=0}^{k}\binom{k}s \big(n^k - (\frac{k(k-1)}2+s^2) n^{k-1}\big) + O(n^{k-2})\\
&=\frac{2^k}{k!}n^k - \frac{2^{k-2} (3k-1)}{(k-1)!}n^{k-1}  + O(n^{k-2}).
\end{split}
Since
$$\binom{2n-(\beta\pm\epsilon)k}{k} = \frac{2^k}{k!}n^k - \frac{2^{k-2}((2(\beta\pm\epsilon)+1)k-1)}{(k-1)!}n^{k-1} + O(n^{k-2}),$$
it follows with necessity that $\beta=1$. Then from
$$S(n,k) - \binom{2n-(1\pm\epsilon)k}{k} = \pm \epsilon\frac{2^{k-1} k}{(k-1)!}n^{k-1} + O(n^{k-2})$$
it follows that the proposed bounds do hold for large enough $n$.

P.S. Just in case, $S(n,k)$ has the following generating function:
$$\sum_{n,k\geq 0} S(n,k)z^n t^k = \frac1{1-(1+t)z-t(1+t)z^2}.$$
