Simplifying a combinatorial expression $\sum\limits_{i=1}^{k-1}i(2k-2-i)\binom{2k}{2i+1}$ Find 
\begin{eqnarray}
\sum_{i=1}^{k-1}i(2k-2-i)\binom{2k}{2i+1}
\end{eqnarray}
I know how to find $\sum_{i=1}^{k-1}a_i\binom{2k}{2i+1}$ if $a_i$ is linear in $i$, but got stuck when $a_i$ is quadratic in $i$. Any idea?
 A: Because you already did the linear case, so I will give you in this answer how can you compute the following sum:
$$S_k(x)=\sum_{0\leq 2i+1\leq k} 2i(2i+1){k \choose 2i+1}x^{2i+1} $$
If we have a formula for $S_k$ then we can compute any quadratic sum (left as exercise). We consider $C_k$ the same way as $S_k$ just in order to sum over all integers (odd + even):
$$C_k(x)=\sum_{0\leq 2i+1\leq k} 2i(2i-1){k \choose 2i+1}x^{2i} $$
and we consider also:
$$f_k(x)=\sum_{0 \leq i \leq k} i(i-1){k \choose i}x^{i} $$
and now we have:
$$C_k(x)+S_k(x)=f_k(x) \ \ \ \ \ \ \ \ \ C_k(x)-S_k(x)=f_k(-x)$$.
So the only things we need to do is to compute $f_k(x)$, and this can be seen , somehow, as the derivative of $(1+x)^k$:
$$ f_k(x)=x^2\sum_{i=0}^{k}i(i-1){k \choose i}x^{k-2}=x^2\frac{d}{d^2x}\left((x+1)^k\right)=k(k-1)x^2(x+1)^{k-2}$$
and hence :
$$S_k(1)=\frac{f_k(1)-f_k(-1)}{2}$$
I hope you can continue from here.
A: One possible method is to use the series
\begin{align}
S_{k}(t) = \sum_{i=0}^{k-1} \binom{2k}{2i+1} \, t^{i} 
\end{align}
to obtain
\begin{align}
\sum_{i=0}^{k-1} \, i \, (2k-2-i) \, \binom{2k}{2i+1} \, t^{i} = - t^{2k-1} \, \frac{d}{dt} \left[ t^{3 - 2k} \, \frac{d \, S_{k}(t)}{dt} \right].
\end{align}
The remaining portion is to determine the value of $S_{k}(t)$ and apply the differentiation and finally let $t \to 1$. 
