Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$

Since I am a Bear of Very Little Brain, and long proofs Bother me, let me post slightly shorter version of essentially the same proof.

$2^q\sum_{k=0}^r(-1)^k\binom rk\binom{k/2}q$ is the coefficient of $z^q$ in the expansion of $(1-\sqrt{1+2z})^r=\left(\frac{-2z}{1+\sqrt{1+2z}}\right)^r$.

Now we want to substitue $\sqrt{1+2z}$ by $1+w$. There is a purely algebraic lemma for this, but one way to establish it is to write this coefficient as a (complex) integral and apply the change of variables formula for integrals: \begin{multline} \DeclareMathOperator{\res}{res} %\res\,(1-\sqrt{1+2z})^r\frac{dz}{z^{q+1}}= \res\left(\frac{-2z}{1+\sqrt{1+2z}}\right)^r\frac{dz}{z^{q+1}}= (-2)^r\res\frac1{(1+\sqrt{1+2z})^r}\frac{dz}{z^{q-r+1}}=\\ (-2)^r\res\frac1{2+w}\frac{dw+w\,dw}{(w+w^2/2)^{q-r+1}}= (-1)^r2^{2r-q-1} \res\frac1{(2+w)^{q-r+1}}\frac{dw+w\,dw}{w^{q-r+1}}. \end{multline} (here $\res_z=\frac1{2\pi i}\oint_{|z|=\epsilon}$, if you will; since $z=w+w^2/2$, $dz=dw+w\,dw$).

So we get a sum of two binomial coefficients (each multiplied by $(-1)^\cdots2^\cdots$) — that's the answer Mathematica gave you.