How to prove the following equation or how to prove when k increases L.H.S also increases for fixed integer r. How to prove $\sum\limits_{i=0}^{\lfloor\frac{r}{2}\rfloor}\binom{r-\frac{k}{2}}{i}\big(\sum\limits_{l=0}^{r-2i}\binom{r-\frac{k}{2}-i}{l}2^l\binom{k}{r-2i-l}\big)=\binom{2r}{r}$ or help me to show when k increases L.H.S also increases for fixed r. I checked in computer search L.H.S is equal to $\binom{2r}{r}$. I tried to prove in the way of markus-scheuer's answer to the question How to prove $\sum\limits_{i=0}^{\lfloor\frac{r}{2}\rfloor}\binom{r}{i}\binom{r-i}{r-2i}2^{r-2i}=\binom{2r}{r}$ by using generating function. What I got is $[v^r](1+v)^k(1+\frac{2}{v})^{r-\frac{k}{2}}(1+\frac{1}{v^2+2v})^{r-\frac{k}{2}}.$ I don't know how to solve after this. Please help me to solve. 
 A: We show that
$$\sum_{q=0}^{\lfloor r/2\rfloor} {r-k\choose q}
\sum_{l=0}^{r-2q} {r-k-q\choose l} 2^l {2k\choose r-2q-l} =
{2r\choose r}.$$
For the inner sum we introduce
$${2k\choose r-2q-l} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-2q-l+1}} (1+z)^{2k}
\; dz.$$
This has the  property that it vanishes when $l\gt  r-2q$ and hence we
may extend the range of $l$ to infinity. We get for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-2q+1}} (1+z)^{2k}
\sum_{l\ge 0} {r-k-q\choose l} 2^l z^l
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-2q+1}} (1+z)^{2k}
(1+2z)^{r-k-q}
\; dz.$$
This too has the convenient property that the pole disappears when
$2q\gt r$ and hence we may extend the range of $q$ to infinity as
well, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r+1}} (1+z)^{2k}
(1+2z)^{r-k}
\sum_{q\ge 0}  {r-k\choose q} \frac{z^{2q}}{(1+2z)^q}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r+1}} (1+z)^{2k}
(1+2z)^{r-k}
\left(1+\frac{z^2}{1+2z}\right)^{r-k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r+1}} (1+z)^{2k}
(1+z)^{2r-2k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r+1}} (1+z)^{2r}
\; dz
\\ = {2r\choose r}.$$
This concludes the argument.
A: It seems there were small typos or a misunderstanding when applying the substitution rule. It should be $2v$ instead of $\frac{2}{v}$ and $\frac{v^2}{1+2v}$ instead of $\frac{1}{v^2+2v}$.

We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^{\lfloor\frac{r}{2}\rfloor}}&\color{blue}{\binom{r-\frac{k}{2}}{i}\sum_{l=0}^{r-2i}\binom{r-\frac{k}{2}-i}{l}2^l\binom{k}{r-2i-l}}\\
&=\sum_{i=0}^\infty [z^i](1+z)^{r-\frac{k}{2}}\sum_{l=0}^\infty [u^l](1+2u)^{r-\frac{k}{2}-i}[v^{r-2i-l}](1+v)^k\tag{1}\\
&=[v^r](1+v)^k\sum_{i=0}^\infty v^{2i}[z^i](1+z)^{r-\frac{k}{2}}\sum_{l=0}^\infty v^l[u^l](1+2u)^{r-\frac{k}{2}-i}\tag{2}\\
&=[v^r](1+v)^k\sum_{i=0}^\infty v^{2i}[z^i](1+z)^{r-\frac{k}{2}}(1+2v)^{r-\frac{k}{2}-i}\tag{3}\\
&=[v^r](1+v)^k(1+2v)^{r-\frac{k}{2}}\sum_{i=0}^\infty \left(\frac{v^2}{1+2v}\right)^i[z^i](1+z)^{r-\frac{k}{2}}\tag{4}\\
&=[v^r](1+v)^k(1+2v)^{r-\frac{k}{2}} \left(1+\frac{v^2}{1+2v}\right)^{r-\frac{k}{2}}\tag{5}\\
&=[v^r](1+v)^{2r}\tag{6}\\
&\color{blue}{=\binom{2r}{r}}\tag{7}
\end{align*}

Comment:


*

*In (1) we apply the coefficient of operator three times and set the upper limit of the sums to $\infty$ without changing anything, since we are adding zeros only.

*In (2) we use the linearity of the coefficient of operator, do some rearrangements and use the rule
\begin{align*}
[u^{p-q}]A(u)=[u^p]u^qA(u)
\end{align*}

*In (3) we apply the substitution rule of the coefficient of operator with $u:=v$ to the inner sum 
\begin{align*}
A(v)=\sum_{i=0}^\infty a_i v^i=\sum_{i=0}^\infty v^i [u^i]A(u)
\end{align*}

*In (4) we do some rearrangements as preparation for the next substitution.

*In (5) we apply the substitution rule of the coefficient of operator with $z:=\frac{v^2}{1+2v}$.

*In (6) we do some simplifications.

*In (7) we select the coefficient of $v^{r}$.
