Showing $\binom{2d-1}{n}-(n+1){d-1\choose n}=\sum_{i=1}^n (-1)^{n-i}{n+1\choose i+1}{id-d+n\choose n}$ for $n\geq1$ and $d\geq n+1$

Question

Recently, when I was trying to compute Hodge numbers of hypersurfaces in toric varieties, I discovered the following combinatorial identity:

For every positive integers $n\geq 1$ and $d\geq n+1$ the following is an identity $$ \binom{2d-1}{n}-(n+1){d-1\choose n}=\sum_{i=1}^n (-1)^{n-i}{n+1\choose i+1}{id-d+n\choose n} $$

I tried to prove it by induction on $n$; for $n=1$ the identity becomes trivial, but the inductive step seems to be very hard (at least for me!).

For $d=n+1$ the identity can be proved by using arguments of V. Batyrev involving polar duality of reflexive polytopes. But for greater $d$ I have not any idea...

Someone can help me? Thank you a lot!

Answer 1

It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g. \begin{align*} \binom{n}{k}=[z^k](1+z)^n\tag{1} \end{align*}

We start with the right-hand side and obtain for $d-1\geq n\geq 1$: \begin{align*} \color{blue}{\sum_{i=1}^n}&\color{blue}{(-1)^{n-i}\binom{n+1}{i+1}\binom{id-d+n}{n}}\\ &=\sum_{i=2}^{n+1}(-1)^{n+1-i}\binom{n+1}{i}\binom{id-2d+n}{n}\tag{2}\\ &=\sum_{i=2}^{n+1}(-1)^{n+1-i}\binom{n+1}{i}[z^n](1+z)^{id-2d+n}\tag{3}\\ &=[z^n](1+z)^{n-2d}\left((1+z)^d-1\right)^{n+1}\\ &\qquad-(-1)^{n+1}[z^n](1+z)^{n-2d}\left(1-(n+1)(1+z)^d\right)\tag{4}\\ &=(-1)^n[z^n](1+z)^{n-2d}-(-1)^n(n+1)(1+z)^{n-d}\tag{5}\\ &=(-1)^n\binom{n-2d}{n}-(-1)^n(n+1)\binom{n-d}{n}\tag{6}\\ &\,\,\color{blue}{=\binom{2d-1}{n}-(n+1)\binom{d-1}{n}}\tag{7} \end{align*} and the claim follows.

Comment:

• In (2) we shift the index and start with $i=2$.

• In (3) we use the coefficient of operator according to (1).

• In (4) we apply the binomial theorem and subtract the terms with $i=0$ and $i=1$ as compensation.

• In (5) we observe that $((1+z)^d-1)^{n+1}=\left(dz+\binom{d}{2}z^2+\cdots\right)^{n+1}$ is a polynomial in $z$ with powers greater than $n$. So, the coefficient of $z^n$ of the left-most term is zero.

• In (6) we select the coefficient of $z^n$.

• In (7) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ twice.

Answer 2

First note that as an order $n+1$ repeated difference of a degree $n$ polynomial, $$ \overbrace{\sum_{i=-1}^n(-1)^{n-i}\binom{n+1}{i+1}}^\text{order $n+1$ difference}\overbrace{\binom{id-d+n}{n}\vphantom{\sum_{i=-1}^n}}^\text{degree $n$ polynomial}=0\tag1 $$ The difference between $(1)$ and the sum in the question are the $i=-1$ and $i=0$ terms. Thus, $$ \begin{align} \sum_{i=1}^n(-1)^{n-i}\binom{n+1}{i+1}\binom{id-d+n}{n} &=\textstyle\overbrace{\color{#C00}{(-1)^n\binom{n+1}{0}\binom{-2d+n}{n}}}^{\text{negative of $i=-1$ term of $(1)$}}-\overbrace{\color{#090}{(-1)^n\binom{n+1}{1}\binom{-d+n}{n}}}^{\text{$i=0$ term of $(1)$}}\\[3pt] &=\color{#C00}{\binom{2d-1}{n}}-\color{#090}{(n+1)\binom{d-1}{n}}\tag2 \end{align} $$ where in the last step of $(2)$, we evaluated the negative binomial coefficients.