Sum of terms involving binomial coefficients and floor functions I would like to know how to prove the following identity.
Let m be a positive integer. Then 
$$\sum_{i=0}^{m+1} (-1)^i \Bigl\lfloor \frac{i}{2} \Bigr\rfloor {2m+1 \choose m+i}=2^{2m-2}$$
Here $\Bigl\lfloor \frac{i}{2} \Bigr\rfloor$ denotes the floor function.
A computational proof is welcome. On the other hand, is it possible to give a combinatorial proof? For example, we interpret the left-hand side as coefficients of a variable in a polynomial.

We seek to evaluate
$$\sum_{q=0}^{m+1} (-1)^q
\Big{\lfloor} \frac{q}{2} \Big{\rfloor} {2m+1\choose m+q}
= \sum_{q=0}^{m+1} (-1)^q
\Big{\lfloor} \frac{q}{2} \Big{\rfloor}
{2m+1\choose m+1-q}.$$
Observe that
$$\Big{\lfloor} \frac{q}{2} \Big{\rfloor}
= [z^q] \left(\sum_{p\ge 0} p z^{2p}
+ \sum_{p\ge 0} p z^{2p+1}\right)
= [z^q]  (1+z) \frac{z^2}{(1-z^2)^2}
\\ = [z^q] \frac{z^2}{(1-z)^2 (1+z)}.$$
We thus have for the sum
$$\sum_{q=0}^{m+1} (-1)^q [z^q] \frac{z^2}{(1-z)^2 (1+z)}
[w^{m+1-q}] (1+w)^{2m+1}
\\ = [w^{m+1}] (1+w)^{2m+1}
\sum_{q=0}^{m+1} w^q (-1)^q [z^q] \frac{z^2}{(1-z)^2 (1+z)}.$$
Now when $q\gt m+1$ the sum term does not contribute to the coefficient
extractor $[w^{m+1}]$ and hence we may extend $q$ to infinity,
getting
$$[w^{m+1}] (1+w)^{2m+1}
\sum_{q\ge 0} w^q (-1)^q [z^q] \frac{z^2}{(1-z)^2 (1+z)}.$$
Here  we  may  apply  the substitution  rule  or  alternatively,  an
annihilated coefficient extractor (ACE) to obtain
$$[w^{m+1}] (1+w)^{2m+1}
\frac{w^2}{(1+w)^2 (1-w)}
= [w^{m-1}]\frac{1}{1-w} (1+w)^{2m-1}.$$
This is
$$\sum_{q=0}^{m-1} {2m-1\choose q}
= \frac{1}{2} 2^{2m-1} = 2^{2m-2}$$
as claimed.
