I would like to know how to obtain (if it exists) a closed form expression of the sum
$$S=\sum^{n}_{k=0}2^k{{n+1}\choose k}{{r-n-2}\choose {n-k}}$$
So far, I have tried to use the method of exponential generating functions to tackle this problem, but to no avail. That is,
$$A(z)B(z)=\left(\sum^{\infty}_{n=0}a_n\frac{z^n}{n!}\right)\left(\sum^{\infty}_{n=0}b_n\frac{z^n}{n!}\right)=\sum^{\infty}_{n=0}\left(\sum^{n}_{k=0}{n\choose k}a_kb_{n-k}\right)\frac{z^n}{n!}$$
I then put the sum into the appropriate format.
$$S=(n+1)(r-n-2)!\sum_{k=0}^{n}{n\choose k}\frac{2^k}{(n-k+1)!(r-2n+k-2)!}$$
Leting $a_k=\frac{2^k}{(n-k+1)!}$ and $b_k=\frac{1}{(r-2n+k-2)!}$,
$$A(z)=\sum^{\infty}_{k=0}\frac{(2x)^k}{(n-k+1)!k!}=\frac{1}{(n+1)!}(1+2z)^{n+1}$$
$$B(z)=\sum^{\infty}_{k=0}\frac{x^k}{(r-n-k-2)!k!}=\frac{1}{(r-n-2)!}(1+z)^{r-n-2}$$
So
$$S=[z^n](1+2z)^{n+1}(1+z)^{r-n-2}$$
and I am stuck. If I let $a_k=\frac{(\sqrt2)^k}{(n-k+1)!}$ and $b_k=\frac{(\sqrt2)^k}{(r-2n+k-2)!}$ I get
$$S=[z^n](1+\sqrt2x)^{r-1}={{r-1}\choose n}2^{\frac{n}{2}}$$
but this answer doesn't even make sense since it isn't even an integer.
I have no idea how to evaluate this sum, and would greatly appreciate any help offered.
Probably not the answer you want but: $$\frac{((n+1)\cdot\Gamma(r-1-n)\cdot _2F_1({-(n+1), -n};{-2n+r-1};2))}{(\Gamma(n+2) \cdot \Gamma(r-1-2n))}$$ Can be derived or using the reduce zeilberg system, which is what I used, but the manual derivation is not hard if wanted. You can also verify by evaluating the the hypergeometric terms
Hyper terms->pochhammer terms->gamma terms->factorial terms