# Adding two Summations (Leibniz formula)

I am wondering how this development is made. I would be really greatful if anyone would like to explain that to me. I don't understand why the right side wont have the form of the product rule like you could see the left side has. This is all from a proof of Leibniz formula by the way.

\begin{align} &\sum_{k=0}^{p-1}{p-1 \choose k}{D^{p-k}fD^{k}g + \sum_{l=1}^{p}{p-1 \choose l-1}{D^{p-l}fD^{l}g} =\\{D^{p}fg} +\sum_{k=1}^{p-1}[{p-1 \choose k}+{p-1 \choose k-1}]{D^{p-k}fD^{k}g+f{D^{p}g}}}. \end{align}

Thank you!

-
I think there are missing brackets around the sum of binomial coefficients in the right hand side. Look at the boundaries in the summation. – k.stm Nov 1 '12 at 16:17
Yes that is true! I will add the brackets. Thank you for pointing that out. – Lukas Arvidsson Nov 1 '12 at 16:43

It looks like all the step is amounts to separating off the $k=0$ term of the first sum on the left, and the $l=p$ term in the second sum. These are your $D^pfg$ and $fD^pg$ terms, and the rest of both sums now have summation range from 1 to $p-1$. In that range it doesn't matter if we replace index $l$ by index $k$.