I am trying to prove that for natural numbers $m \geq 1$ and $1 < l \leq m$, the following binomial identity is true:
$$ \sum_{j=1}^m 2^{2m-2j} (-1)^{j+l} \binom{2m-j}{2m-2j} \binom{j}{l} = \binom{2m+1}{2l+1}.$$
I have tried to induct on $m$, using Pascal's identity, but then I am left with a term of the form $$ \sum_{j=1}^m 2^{2m-2j} (-1)^{j+l} \binom{2m-j-1}{2m-2j} \binom{j}{l},$$
that I cannot evaluate. I have also considered induction on $l$ (possible abuse of terminology), using the identity:
$$ \binom{n}{k} = \frac{n-k+1}{k} \binom{n}{k-1}.$$
In this case, we have to deal with the sum
$$ \sum_{j=1}^m 2^{2m-2j} (-1)^{j+l} j \binom{2m-j-1}{2m-2j} \binom{j}{l-1},$$
which I have not been able to express in a closed form. I have also attempted to think about this combinatorially, though not for very long, and I could not visualise how this summation counted the number of subsets of size $2l+1$ in a set of size $2m+1$.
If anyone could point me towards a proof of this identity, I would be very grateful. I would prefer a hint, e.g. an identity that I am missing, rather than a full solution if possible. It might also be that the identity I am trying to prove is false. In this case, please let me know as soon as possible!
