TITLE: summing this binomial series
QUESTION [6 upvotes]: I found a really interesting question which is as follows:
Prove that the value of
$$\sum^{7}_{k=0}[({7\choose k}/{14\choose k})*\sum^{14}_{r=k}{r\choose k}{14\choose r}] = 6^7$$
my approach:
I tried to simplify the innermost sigma as well as  trying to simplify by using
${n\choose k}=n!/k!(n-k)!$ however I am can't get hold of this one.
My guess is that the summation simplifies into a standard series but I can't say for sure.
Kindly help me out.

REPLY [4 votes]: Setting $n=7$ we obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{n}{k}\binom{2n}{k}^{-1}\sum_{r=k}^{2n}\binom{r}{k}\binom{2n}{r}}\\
&=\sum_{k=0}^n\binom{n}{k}\frac{k!(2n-k)!}{(2n)!}\sum_{r=k}^{2n}\frac{r!}{k!(r-k)!}\,\frac{(2n)!}{r!(2n-r)!}\\
&=\sum_{k=0}^n\binom{n}{k}\sum_{r=k}^{2n}\binom{2n-k}{r-k}\\
&=\sum_{k=0}^n\binom{n }{k}\sum_{r=0}^{2n-k}\binom{2n-k}{r}\\
&=\sum_{k=0}^n\binom{n}{k}2^{2n-k}\\
&=2^{2n}\sum_{k=0}^n\binom{n}{k}\frac{1}{2^k}\\
&=2^{2n}\left(1+\frac{1}{2}\right)^n\\
&\,\,\color{blue}{=6^n}
\end{align*}
and the claim follows.