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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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    It think you should use ${n \choose k}={{n-1} \choose k}+{{n-1} \choose {k-1}}$ to telescope.2012-04-10

1 Answers 1

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Let’s replace $n+1$ by $n$ in order to simplify the notation; the conjectured identity then becomes

$\binom{d-1}n=\sum_{i=0}^n(-1)^{n+i}\binom{d}i=(-1)^n\sum_{i=0}^n(-1)^i\binom{d}i\;.$

Suppose first that $d\le n$. Clearly $\dbinom{d-1}n=0$, and

$\sum_{i=0}^n(-1)^i\binom{d}i=\sum_{i=0}^d(-1)^i\binom{d}i=(1-1)^d=0$ by the binomial theorem, since $d\ge 1$.

Now suppose that $1\le n. Then

$\begin{align*} \sum_{i=0}^n(-1)^i\binom{d}i&=\sum_{i=0}^n(-1)^i\left(\binom{d-1}i+\binom{d-1}{i-1}\right)\\ &=\sum_{i=0}^n(-1)^i\binom{d-1}i+\sum_{i=1}^n(-1)^i\binom{d-1}{i-1}\\ &=\sum_{i=0}^n(-1)^i\binom{d-1}i+\sum_{i=0}^{n-1}(-1)^{i+1}\binom{d-1}i\\ &=\sum_{i=0}^n(-1)^i\binom{d-1}i-\sum_{i=0}^{n-1}(-1)^i\binom{d-1}i\\ &=(-1)^n\binom{d-1}n\;, \end{align*}$

and the identity follows immediately by multiplying by $(-1)^n$.

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    @Brain I had this in mind, but since I never deal with this type of problems, I flinched!2012-04-10