Half of Vandermonde's Identity We know Vandermonde's Identity, which states 
$\sum_{k=0}^{r}{m \choose k}{n \choose r-k}={m+n \choose r}$.
Does anyone know what happens if we walk bigger steps with k? Say we skip all the odd ks, is something like
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\frac{1}{2} {m+n \choose r}$
or at least 
$\sum_{k=0}^{r/2}{m \choose 2k}{n \choose r-2k}=\Theta \left( \frac{1}{2} {m+n \choose r}\right)$
true? 
Maybe someone here has even some general insight on other step widths?
Thank you!
 A: We derive a binomial identity which shows the deviation of OPs sum from $\frac{1}{2}\binom{m+n}{r}$. It  is convenient to use the  coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}

We assume wlog $n\geq m$ and obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^{r/2}}&\color{blue}{\binom{m}{2k}\binom{n}{r-2k}}\\
&=\sum_{k\geq 0}\binom{m}{2k}[z^{r-2k}](1+z)^n\tag{2}\\
&=[z^r](1+z)^n\sum_{k\geq 0}\binom{m}{2k}z^{2k}\tag{3}\\
&=[z^r](1+z)^n\frac{1}{2}\left((1+z)^m+(1-z)^m\right)\tag{4}\\
&=\frac{1}{2}[z^r](1+z)^{m+n}+\frac{1}{2}[z^r](1+z)^n(1-z)^m\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}[z^r](1-z^2)^m(1+z)^{n-m}\tag{5}\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}[z^r]\sum_{k=0}^{r/2}\binom{m}{k}(-1)^kz^{2k}(1+z)^{n-m}\\
&=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}\sum_{k=0}^{r/2}\binom{m}{k}(-1)^k[z^{r-2k}](1+z)^{n-m}\tag{6}\\
&\,\,\color{blue}{=\frac{1}{2}\binom{m+n}{r}+\frac{1}{2}\sum_{k=0}^{r/2}\binom{m}{k}\binom{n-m}{r-2k}(-1)^k}\tag{7}
\end{align*}

Comment:


*

*In (2) we apply the coefficient of operator as indicated in (1) and we set the upper limit of the sum to $\infty$ without changing anything since we are adding zeros only.

*In (3) we use the linearity of the coefficient of operator.

*In (4) we write the sum as polynomial in closed form.

*In (5) we select the coefficient of $z^r$ of the left polynomial and we rewrite the other polynomial keeping in mind that $n\geq m$.

*In (6) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (7) we select the coefficient of $z^{r-2k}$.
A: In general, having the ogf (z-Transform)
$$
F(z) = \sum\limits_{0\, \le \;n} {a_{\,n} \,z^{\,n} } 
$$
then
$$
{1 \over m}\sum\limits_{0 \le \,k\, \le \,m - 1} {\left( {z^{\,{1 \over m}} \;e^{\,i\,{{2k\pi } \over m}} } \right)^{\,j}
 F(z^{\,{1 \over m}} \;e^{\,i\,{{2k\pi } \over m}} )}
  = \sum\limits_{0\, \le \;n} {\,a_{\,m\;n - j} \,z^{\,n} } 
$$
But unfortunately, the truncated binomial expansion
$$
\sum\limits_{0\, \le \;k} {\left( \matrix{  n \cr   r - k \cr}  \right)\,z^{\,k} } 
$$
does not have in general ($r<n$) a compact closed expression.
We can go either through the Hypergeometric version
$$
\sum\limits_{\left( {0\, \le } \right)\;k\,\left( { \le \,\,r} \right)} {
 \binom{m}{k} \binom{n}{r-k}\,z^{\,k} } 
  = \binom{n}{r} \;{}_2F_{\,1} \left( {\matrix{
   { - m,\; - r}  \cr 
   {n - r + 1}  \cr 
 } \;\left| {\,z} \right.} \right)
$$
or through the double ogf
$$
\eqalign{
  & G(x,y,n,m) = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {
  \binom{m}{j}\,\binom{n}{k-j} y^{\,j} } } \right)x^{\,k} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {
  \binom{m}{j}\left( {x\,y} \right)^{\,j} \sum\limits_{\left( {j\, \le } \right)\,k\,\left( { \le \,n} \right)\,} {  \,\binom{n}{k-j}x^{\,k - j} } }  =   \cr 
  &  = \left( {1 + xy} \right)^{\,m} \left( {1 + x} \right)^{\,n}  \cr} 
$$
Then for instance we have
$$
\eqalign{
  & \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,\left\lfloor {\min (m,k)/2} \right\rfloor } \right)\;} {
 \left( \matrix{ m \cr   2j \cr}  \right)\,\left( \matrix{  n \cr   k - 2j \cr}  \right)} } \right)x^{\,k} }  =   \cr 
  &  = {1 \over 2}\left( {G(x,1,n,m) + G(x, - 1,n,m)} \right) =   \cr 
  &  = {1 \over 2}\left( {1 + x} \right)^{\,n} \left( {\left( {1 + x} \right)^{\,m}  + \left( {1 - x} \right)^{\,m} } \right) =   \cr 
  &  = {1 \over 2}\left( {1 + x} \right)^{\,n + m}  + {1 \over 2}\left( {1 + x} \right)^{\,n} \left( {1 - x} \right)^{\,m}  =   \cr 
  &  = {1 \over 2}\left( {1 + x} \right)^{\,n + m}  + {1 \over 2}\left( {1 + x} \right)^{\,n - m} \left( {1 - x^{\,2} } \right)^{\,m}  =   \cr 
  &  = {1 \over 2}\left( {1 + x} \right)^{\,n + m}  + {1 \over 2}\left( {1 - x^{\,2} } \right)^{\,{{n + m} \over 2}}
 \left( {{{1 + x} \over {1 - x}}} \right)^{\,{{n - m} \over 2}}  \cr} 
$$
which clearly indicates what is the difference between
$$
{1 \over 2}\binom{n+m}{r}
\quad vs\quad \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,\left\lfloor {\min (m,k)/2} \right\rfloor } \right)\;} {
\binom{m}{2j} \, \binom{n}{r-2j} }
$$
Of course the complement will be
$$
\eqalign{
  & \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,\left\lfloor {\min (m,k)/2} \right\rfloor } \right)\;}  {
\binom{m}{2j+1} \,\binom{n}{k - \left( {2j + 1} \right)}} } \right)x^{\,k} }  =   \cr 
  &  = {1 \over 2}\left( {G(x,1,n,m) - G(x, - 1,n,m)} \right) =   \cr 
  &  = {1 \over 2}\left( {1 + x} \right)^{\,n} \left( {\left( {1 + x} \right)^{\,m}  - \left( {1 - x} \right)^{\,m} } \right) \cr} 
$$
