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.
 A: For $i\in\mathbb{Z}$,
$$
\left\lfloor\frac i2\right\rfloor=\frac i2-\frac14+\frac{(-1)^i}4\tag1
$$
Furthermore,
$$
(-1)^i\binom{i}{1}=(-1)^i\binom{i}{i-1}=-\binom{-2}{i-1}\tag2
$$
and
$$
(-1)^i=\binom{-1}{i}\tag3
$$
Thus,
$$
\begin{align}
&\sum_{i=0}^{m+1}(-1)^i\left\lfloor\frac i2\right\rfloor\binom{2m+1}{m+i}\tag4\\
&=\sum_{i=0}^{m+1}(-1)^i\left(\frac i2-\frac14+\frac{(-1)^i}4\right)\binom{2m+1}{m+i}\tag5\\
&=\color{#C00}{-\frac12\sum_{i=0}^{m+1}\binom{-2}{i-1}\binom{2m+1}{m-i+1}}\color{#090}{-\frac14\sum_{i=0}^{m+1}\binom{-1}{i}\binom{2m+1}{m-i+1}}\\
&\color{#00F}{+\frac14\sum_{i=0}^{m+1}\binom{2m+1}{m+i}}\tag6\\
&=\color{#C00}{-\frac12\binom{2m-1}{m}}\color{#090}{-\frac14\binom{2m}{m+1}}\color{#00F}{+\frac14\binom{2m+1}{m}}\vphantom{\sum_{i=1}^{m+1}}\color{#00F}{+\frac14\sum_{i=1}^{m+1}\binom{2m+1}{m+i}}\tag7\\
&=\color{#C00}{-\frac14\binom{2m}{m}}\color{#090}{-\frac14\binom{2m}{m-1}}\color{#00F}{+\frac14\binom{2m+1}{m}}\vphantom{\sum_{i=1}^{m+1}}\color{#00F}{+\frac18\sum_{i=0}^{2m+1}\binom{2m+1}{i}}\tag8\\[9pt]
&=2^{2m-2}\tag{9}
\end{align}
$$
Explanation:
$(5)$: apply $(1)$
$(6)$: apply $(2)$ and $(3)$
$(7)$: apply Vandermonde's Identity
$(8)$: $\overbrace{\binom{2m-1}{m-1}+\binom{2m-1}{m}}^{\text{equal by symmetry}}=\binom{2m}{m}$ and symmetry yields $\sum\limits_{i=1}^{m+1}\binom{2m+1}{m+i}=\frac12\sum\limits_{i=0}^{2m+1}\binom{2m+1}{i}$
$(9)$: $\binom{2m}{m}+\binom{2m}{m-1}=\binom{2m+1}{m}$ and $\sum\limits_{i=0}^{2m+1}\binom{2m+1}{i}=2^{2m+1}$
A: 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.
