The following (conjectured) identity has come up in a research problem that I am working on:

for even $a$ $$\sum_{i=0}^{a-1} (-1)^{a-i}\binom{a}{i} \binom{2m-i-2}{m-i-1}=0;$$ and for odd $a$ $$\sum_{i=0}^{a-1} (-1)^{a-i}\binom{a}{i} \binom{2m-i-2}{m-i-1}=-2\binom{2m-a-2}{m-a-1},$$

where $a,m$ are positive integers with $1\le a\le m-2$.

I've verified the identity holds for small values of $a,m$.

The closest problem I have found is Help with a Binomial Coefficient Identity. Any suggestion how to apply that identity or to find another proof?


If you extend the sum to $a$, you can combine the even and odd cases into

$$ \sum_{i=0}^a(-1)^i\binom ai\binom{2n-i}{n-i}=\binom{2n-a}n\;, $$

with $n=m-1$. This is a double count using inclusion–exclusion of the number of ways of selecting $n$ from $2n$ elements such that $a$ particular elements are not included in the selection.


For an algebraic proof of the re-formulated identity by @joriki we write

$$\sum_{q=0}^a (-1)^q {a\choose q} {2n-q\choose n-q} = \sum_{q=0}^a (-1)^q {a\choose q} [z^{n-q}] (1+z)^{2n-q} \\ = [z^n] (1+z)^{2n} \sum_{q=0}^a (-1)^q {a\choose q} z^q (1+z)^{-q} \\ = [z^n] (1+z)^{2n} \left(1-\frac{z}{1+z}\right)^a = [z^n] (1+z)^{2n-a} = {2n-a\choose n}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.