TITLE: How to prove $\sum\limits_{i=0}^{\lfloor\frac{r}{2}\rfloor}\binom{r}{i}\binom{r-i}{r-2i}2^{r-2i}=\binom{2r}{r}$
QUESTION [6 upvotes]: Please help me to prove $\sum\limits_{i=0}^{\lfloor r/2\rfloor}\binom{r}{i}\binom{r-i}{r-2i}2^{r-2i}=\binom{2r}{r}$. By computer search I have found these for r varies from 0 to 10000. How to prove this for general $r\in N.$

REPLY [7 votes]: Here is a proof based upon generating functions. It is convenient to use the coefficient of operator $[z^i]$ to denote the coefficient of $z^i$ in a series. This way we can write e.g.
\begin{align*}
[z^i](1+z)^n=\binom{n}{i}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^{\lfloor r/2\rfloor}}&\color{blue}{\binom{r}{i}\binom{r-i}{r-2i}2^{r-2i}}\\
&=\sum_{i=0}^\infty[z^{i}](1+z)^{r}[u^{r-2i}](1+2u)^{r-i}\tag{1}\\
&=[u^r](1+2u)^r\sum_{i=0}^\infty\left(\frac{u^2}{1+2u}\right)^i[z^i](1+z)^r\tag{2}\\
&=[u^r](1+2u)^r\left(1+\frac{u^2}{1+2u}\right)^r\tag{3}\\
&=[u^r](1+u)^{2r}\tag{4}\\
&\color{blue}{=\binom{2r}{r}}\tag{5}
\end{align*}
  and the claim follows.

Comment:

In (1) we apply the coefficient of operator twice and set the upper limit of the sum 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 $z:=\frac{u^2}{1+2u}$
\begin{align*}
A(u)=\sum_{i=0}^\infty a_i u^i=\sum_{i=0}^\infty u^i [z^i]A(z)
\end{align*}
In (4) we do some simplifications.
In (5) we select the coefficient of $u^r$.