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?
 A: 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!
A: By way  of enrichment  and presenting various  techniques we  start by
introducting
$$S_{a,b}(n) = \sum_{q=0}^n {a+q\choose b+q} q$$
where $a\ge b$ and write with an Iverson bracket
$$\sum_{q\ge 0} {a+q\choose b+q} q
[[0\le q\le n]]
= \sum_{q\ge 0} {a+q\choose a-b} q
[z^n] \frac{z^q}{1-z}
\\ = [z^n] \frac{1}{1-z}
\sum_{q\ge 0} {a+q\choose a-b} q z^q
= [z^n] \frac{1}{1-z} [w^{a-b}] (1+w)^a
\sum_{q\ge 0} q z^q (1+w)^q
\\ = [z^n] \frac{1}{1-z} [w^{a-b}] (1+w)^a
\frac{z(1+w)}{(1-z(1+w))^2}
= [z^{n-1}] \frac{1}{1-z} [w^{a-b}] (1+w)^{a+1}
\frac{1}{(1-z-zw)^2}
= [z^{n-1}] \frac{1}{(1-z)^3} [w^{a-b}] (1+w)^{a+1}
\frac{1}{(1-zw/(1-z))^2}
\\ = [z^{n-1}] \frac{1}{(1-z)^3}
\sum_{q=0}^{a-b} {a+1\choose a-b-q}
(q+1) \frac{z^q}{(1-z)^q}
\\ = \sum_{q=0}^{a-b} {a+1\choose a-b-q}
(q+1) [z^{n-1-q}] \frac{1}{(1-z)^{q+3}}
\\ = \sum_{q=0}^{a-b} {a+1\choose a-b-q}
(q+1) {n+1\choose q+2}
\\ = (n+1) \sum_{q=0}^{a-b} {a+1\choose a-b-q}
 {n\choose q+1}
- \sum_{q=0}^{a-b} {a+1\choose a-b-q}
{n+1\choose q+2}.$$
The first sum is
$$\sum_{q=0}^{a-b} {a+1\choose a-b-q} {n\choose q+1}
= [z^{a-b}] (1+z)^{a+1}
\sum_{q=0}^{a-b} {n\choose q+1} z^q.$$
We may extend $q$ to infinity  because of the coefficient extractor in
front:
$$[z^{a-b}] (1+z)^{a+1}
\sum_{q\ge 0} {n\choose q+1} z^q
= [z^{a-b+1}] (1+z)^{a+1}
\sum_{q\ge 0} {n\choose q+1} z^{q+1}
\\ = [z^{a-b+1}] (1+z)^{a+1} ((1+z)^n - 1)
= {n+a+1\choose a-b+1} - {a+1\choose a-b+1}.$$
The second is
$$\sum_{q=0}^{a-b} {a+1\choose a-b-q} {n+1\choose q+2}
= [z^{a-b}] (1+z)^{a+1}
\sum_{q=0}^{a-b} {n+1\choose q+2} z^q.$$
We may  once more extend  $q$ to  infinity because of  the coefficient
extractor in front:
$$[z^{a-b}] (1+z)^{a+1}
\sum_{q\ge 0} {n+1\choose q+2} z^q
= [z^{a-b+2}] (1+z)^{a+1}
\sum_{q\ge 0} {n+1\choose q+2} z^{q+2}
\\ = [z^{a-b+2}] (1+z)^{a+1} ((1+z)^{n+1} - 1 - (n+1)z)
\\ = {n+a+2\choose a-b+2} - {a+1\choose a-b+2}
- (n+1) {a+1\choose a-b+1}.$$
Collecting and canceling we obtain at last
$$\bbox[5px,border:2px solid #00A000]{
S_{a,b}(n) = (n+1) {n+a+1\choose a-b+1}
- {n+a+2\choose a-b+2}
+ {a+1\choose a-b+2}}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
S_{a,b}(n) = (n+1) {n+a+1\choose n+b}
- {n+a+2\choose n+b}
+ {a+1\choose b-1}.}$$
This simplifies to the accepted answer.
