How to find a closed form for $\sum_{i=0}^n \binom{a+i}{b+i}i$ Wolframalpha tells me it's $$\frac{b (b + 1) \binom{a + 1}{ b + 1} - (b + n + 1) (b (n + 1) - (a + 1) n) \binom{a + n + 1}{ b + n + 1}}{(a - b + 1) (a - b + 2)}$$ but how to come up with or at least prove that?

This is not quite the same expression, but this is how I would go about finding the closed formula. We can first rewrite the summation as
$$\sum_{i=0}^n \binom{a+i}{b+i}i = \sum_{k=1}^n \sum_{i=k}^n \binom{a+i}{b+i}$$
Then we can use the hockey stick formula to get that the inner sum is
$$\sum_{k=1}^n \binom{a+n+1}{b+n} - \binom{a + k}{b+k-1}$$
The first term is the same every time, so we can rewrite this as
$$n\binom{a+n+1}{b+n} - \sum_{k=1}^n \binom{a+k}{b+k-1}$$
And use the hockey stick formula again to get that this is
$$n\binom{a+n+1}{b+n} - \binom{a+n+1}{b+n-1} + \binom{a+1}{b-1}$$
And there we have a closed formula!
