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Concerns about the arithmetic genus of projective hypersurfaces led me to make the following combinatorial conjecture: $${d-1\choose n+1} =\sum_{i=0}^{n+1} (-1)^{n+i+1} {d\choose i}$$ for $d \geq 1$, $n \geq 0$. If I made no mistakes in my code, I was also able to find some reasonably strong numerical evidence that this is, in fact, an identity. Unfortunately, my skill at combinatorics is sufficiently poor that while I might be able to prove the statement with a fair amount of effort, I doubt I could find an enlightening proof.

Can someone supply an enlightening proof of the statement above?

Obviously, a counterexample would also suffice, but I doubt there is one.

Additional note: I did not see a reasonable way to search for this specific identity, so it is quite possible this is an exact duplicate of another question, in which case my question should be closed.

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    Looks very similar to [equations 11 & 12](http://en.wikipedia.org/wiki/Binomial_coefficient#Series_involving_binomial_coefficients): $$\sum_{j=0}^k(-1)^j{n\choose j}=(-1)^k{n-1\choose k}$$ which is said to hold for all $n\ne0$.2012-04-10
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    @bgins: I think it is in fact the same identity, with my $d$ corresponding to $n$ and my $n$ corresponding to $k-1$. Unfortunately, the suggested proof is "by induction." But it is very helpful to know this is, in fact, a standard identity.2012-04-10
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    Note that they go on to say it's a special case of a theorem about finite differences of a polynomial (equation 12): $$\deg P(x)\le n\implies\sum_{j=0}^n(-1)^j{n\choose j}P(j)=0$$ This is certainly a standard theorem.2012-04-10
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    It think you should use ${n \choose k}={{n-1} \choose k}+{{n-1} \choose {k-1}}$ to telescope.2012-04-10

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