The sum $$S_{n,m} =\sum_{k=0}^{n}~ (-1)^k{n \choose k} {2n-2k\choose n+m-1}$$ can be evaluated by Mathematica for $m=0,1,2,3$ in simple closed forms. But in general $$ S_{n,m}=2^{n-m+1}{n \choose m-1}~~~(*).$$ The question is how to prove (*) by hand?
Prove: $\sum_{k=0}^{n}~(-1)^k{n \choose k} {2n-2k \choose n+m-1}=2^{n-m+1} {n \choose m-1}$
binomial theorembinomial-coefficientssequences-and-seriessummation
Best Answer
We have
$$\sum_{k=0}^n (-1)^k {n\choose k} {2n-2k\choose n+m-1} = [z^{n+m-1}] (1+z)^{2n} \sum_{k=0}^n (-1)^k {n\choose k} (1+z)^{-2k} \\ = [z^{n+m-1}] (1+z)^{2n} \left(1-\frac{1}{(1+z)^2}\right)^n \\ = [z^{n+m-1}] ((1+z)^2-1)^{n} = [z^{n+m+1}] (2z+z^2)^{n} \\= [z^{n+m-1}] z^{n} (z+2)^{n} = [z^{m-1}] (z+2)^n = {n\choose m-1} \times 2^{n-m+1}.$$
The last coefficient extractor returns zero when $m-1\lt 0,$ as does the binomial coefficient.