Error in a binomial coefficient sum identity proof I have been given an identity $\begin{align}
\sum_{i=0}^n i\binom{n}{i}^2 = n\binom{2n-1}{n-1}
\end{align}$.
However when I tried to prove it, I got a different result.
$\begin{align}
\sum_{i=0}^n i\binom{n}{i}^2 = n\sum_{i=0}^n \binom{n-1}{i-1} \binom{n}{i} = 
n\sum_{i=0}^n \binom{n-1}{n-i} \binom{n}{i} = n\binom{2n-1}{n}
\end{align}$
First equation follows from absorption identity, second one from symmetry, and the third one from Vandermonde's identity. Where is my mistake?
 A: Let us devise a purely combinatorial approach: assume to have a parliament with $n$ people in the right wing, $n$ people in the left wing. In how many ways can we form a committee with $n$ people and elect a chief of the committee from the left wing? The first approach is to select $i$ people from the left wing, $n-i$ people from the right wing, then the chief among the selected $i$ people from the left wing. This leads to $\sum_{i=0}^{n}i\binom{n}{i}\binom{n}{n-i}=\sum_{i=0}^{n}i\binom{n}{i}^2$. The other approach is to select the chief from the left wing first ($n$ ways for doing that), then select $n-1$ people from the remaining $2n-1$ in the parliament. Conclusion:
$$ \sum_{i=0}^{n}i\binom{n}{i}^2 = n\binom{2n-1}{n-1} = n\binom{2n-1}{n}. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{i = 0}^{n}i{n \choose i}^{2} & =
\sum_{i = 0}^{n}i{n \choose i}{n \choose n - i} =
\sum_{i = 0}^{n}i{n \choose i}\bracks{z^{n - i}}\pars{1 + z}^{n} =
\bracks{z^{n}}\pars{1 + z}^{n}\sum_{i = 0}^{n}{n \choose i}i\,z^{i}
\\[5mm] & =
\bracks{z^{n}}\pars{1 + z}^{n}
\bracks{z\,\partiald{}{z}\sum_{i = 0}^{n}{n \choose i}z^{i}} =
\bracks{z^{n}}\pars{1 + z}^{n}\bracks{z\,\partiald{\pars{1 + z}^{n}}{z}}
\\[5mm] & =
\bracks{z^{n - 1}}\pars{1 + z}^{n}\bracks{n\pars{1 + z}^{n - 1}} =
n\bracks{z^{n - 1}}\pars{1 + z}^{2n - 1} =
\bbx{n{2n - 1 \choose n - 1}}
\end{align}
A: If one takes $n=2$ then
$$\sum_{i=0}^n i\binom{n}i^2=0+\binom21^2+2\binom22^2=4+2=6$$
and
$$\binom{2n-1}{n-1}=\binom{3}1=3.$$
It seems in this case that
$\sum_{i=0}^n i\binom{n}i^2$ equals $n\binom{2n-1}n$ rather than
$\binom{2n-1}{n-1}$.
A: The question as stated is incorrect; here's a combinatorial proof for your version (the $n\binom{2n-1}{n}$ version):
Suppose we have $n$ turtles and $n$ rabbits at a sports meet. How many ways can we choose two teams of $n$ animals each, one of which has a captain (who must be a turtle)?
The LHS counts this directly by counting the number of ways to choose $i$ turtles to include and $i$ rabbits to exclude from the team with a captain, and picking a captain from among these turtles. The other team is composed of the remaining animals.
The RHS first picks the captain, then chooses $n$ of the other animals to form the other team. The remaining animals join the captain's team. 
