Reduce two summations

Question

Can someone please show me the steps how to proof the following identity? $$\sum _{n=0}^{a-1} \sum _{l=0}^{a-1} \frac{(n+1) (-1)^{n} \binom{a}{n+1} (l+1) (-1)^{l} \binom{a}{l+1} }{l+n+2}=\frac{a}{4 a-2}$$

Answer 1

Observe we have \begin{align} \int^0_{-1} x \left(\sum^{a-1}_{n=0}(n+1)\binom{a}{n+1}x^n\right)^2\ dx = &\ \sum^{a-1}_{n=0}\sum^{a-1}_{l=0}(n+1)(l+1)\binom{a}{n+1}\binom{a}{l+1}\int^0_{-1} x^{n+l+1}\ dx\\ =&\ \sum^{a-1}_{n=0}\sum^{a-1}_{l=0}(n+1)(l+1)\binom{a}{n+1}\binom{a}{l+1} \frac{-(-1)^{n+l+2}}{n+l+2}. \end{align} On the other hand, we see that \begin{align} -\int^0_{-1}x\left(\sum^{a-1}_{n=0}(n+1)\binom{a}{n+1}x^n \right)^2\ dx =&\ -\int^0_{-1}x\left(\frac{d}{dx}\sum^{a}_{n=0}\binom{a}{n}x^n \right)^2\ dx\\ =&\ -\int^0_{-1}x\left(\frac{d}{dx}(1+x)^a \right)^2\ dx\\ =&\ -\int^0_{-1}a^2x(1+x)^{2a-2}\ dx = \frac{a}{4a-2}. \end{align} Hence we have the desired result.