I [asked](https://mathoverflow.net/questions/376606/binomial-coefficients-sum?noredirect=1#comment955008_376606) how to calculate $$\sum\_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$ and got amazing answers. A bit later, however, I figured I needed something rather more complicated: I need to find the value of $$\sum\_{i = 0}^k(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i},$$ where $k$ can be any integer between $0$ and $b$. Is there any closed formula for that?



An ambitiouser binomial coefficients sum
WolframAlpha claims it is
$$\frac{(-1)^{k + 1} (k + 1) (-a - b + k + 1) \binom{b}{k + 1} \binom{a + b - k - 2}{a - k - 1}}{a b},$$
and you can absorb the $k+1$ and $b$, yielding
$$\frac{(-1)^{k+1} (-a - b + k + 1) \binom{b-1}{k} \binom{a + b - k - 2}{a - k - 1}}{a}.$$
A little simpler:
$$\frac{(-1)^k b \binom{b-1}{k} \binom{a + b - k - 1}{b}}{a}.$$


