An alternating binomial sum equal to 1

Question

I was trying to prove that $$\sum_{j=0}^{r}(-1)^{r-j}\binom{d-j}{r-j}\sum_{l=1}^{d-k}(-1)^{l+1}\binom{d-k}{l}\binom{d-l+1}{j}=1$$ Where $0 \leq r\leq k-1$, $k \leq \lfloor \frac{d}{2} \rfloor$.
I started by reordering the terms like that:

$$(-1)^{r}\sum_{l=1}^{d-k}(-1)^{l+1}\binom{d-k}{l}\sum_{j=0}^{r}(-1)^{j}\binom{d-j}{r-j}\binom{d-l+1}{j}$$ and using one of the variants of Vandermonde convolution, I simplified the inner sum: $$(-1)^{r}\sum_{l=1}^{d-k}(-1)^{l+1}\binom{d-k}{l}\binom{l-1}{r}$$ Does anyone have any suggestions on how I could simplify this last sum to $1?$ I think it might be a known identity...
Thanks everyone for the help.

Answer 1

HINT If your familiar with complex analysis, the Egorychev method gives a nice way to prove this.

Start by writing $\large \binom{l-1}{r} = \frac{1}{2\pi i} \int_\gamma \frac{(1+z)^{l-1}}{z^{r+1}}dz$

Exchange sum and integral and do the sum with aid of the binomial theorem (notice you're missing the term $l=0$)

Use that $\large \frac{1-(-z)^{d-k}}{1-(-z)} = 1 - z + z^2 - z^3 + \dots +(-z)^{d-k-1}$ and the integral you have extracts the $r$th coefficient of this.

Answer 2

Consider $ f:\mathbb{R}^{*}\rightarrow\mathbb{R}, \ x\mapsto\sum\limits_{\ell=1}^{d-k}{\left(-1\right)^{\ell + 1}\binom{d-k}{\ell}x^{\ell - 1}} $.

We have for $ x\in\mathbb{R}^{*} $ : \begin{aligned}f\left(x\right)&=\frac{1}{x}-\frac{1}{x}\sum_{\ell = 0}^{d-k}{\left(-1\right)^{\ell}\binom{d-k}{\ell}x^{\ell}}\\ &=\frac{1}{x}-\frac{\left(1-x\right)^{d-k}}{x}\end{aligned}

Differentiating $ f $, $r $ times gives on one side $ f^{\left(r\right)}\left(x\right)=r!\sum\limits_{\ell=1}^{d-k}{\left(-1\right)^{\ell + 1}\binom{d-k}{\ell}\binom{\ell - 1}{r}x^{\ell - 1-r}} $, and on the other, using leibniz rule : \begin{aligned}f^{\left(r\right)}\left(x\right)&=\frac{\left(-1\right)^{r}r!}{x^{r+1}}-\sum_{p=0}^{r}{\binom{r}{p}\left(x\mapsto\left(1-x\right)^{d-k}\right)^{\left(p\right)}\left(x\right)\left(x\mapsto\frac{1}{x}\right)^{\left(r-p\right)}\left(x\right)} \\&=\frac{\left(-1\right)^{r}r!}{x^{r+1}}-\sum_{p=0}^{r}{\binom{r}{p}\left(-1\right)^{p}\frac{\left(d-k\right)!}{\left(d-k-p\right)!}\left(1-x\right)^{d-k-p}\frac{\left(-1\right)^{r-p}\left(r-p\right)!}{x^{r-p+1}}}\\ &=\frac{\left(-1\right)^{r}r!}{x^{r+1}}\left(1-\sum_{p=0}^{r}{\binom{d-k}{p}x^{p}\left(1-x\right)^{d-k-p}}\right)\end{aligned}

Finally, taking $ x= 1$ gives : $$ f^{\left(r\right)}\left(1\right)=r!\sum\limits_{\ell=1}^{d-k}{\left(-1\right)^{\ell + 1}\binom{d-k}{\ell}\binom{\ell - 1}{r}}=\left(-1\right)^{r}r! $$

Hence the result.

Answer 3

The inner sum is also a Chu-Vandermonde Identity in disguise.

We obtain \begin{align*} {\color{blue}{(-1)^r}}&{\color{blue}{\sum_{l=r+1}^{d-k}\binom{d-k}{l}\binom{l-1}{r}}}\tag{1}\\ &=\sum_{l=r+1}^{d-k}\binom{d-k}{d-k-l}\binom{-r-1}{l-1-r}\tag{2}\\ &=\sum_{l=0}^{d-k-r-1}\binom{d-k}{d-k-r-1-l}\binom{-r-1}{l}\tag{3}\\ &=\binom{d-k-r-1}{d-k-r-1}\tag{4}\\ &\,\,\color{blue}{=1} \end{align*} and the claim follows.

Comment:

• In (1) we start the summation with $l=r+1$, since $\binom{l-1}{r}=0$ if $l<r+1$.

• In (2) we use the binmial identity $\binom{p}{q}=\binom{p}{p-q}$ for both factors and \begin{align*} \binom{-p}{q}=\binom{p+q-1}{q}(-1)^q \end{align*}

• In (3) we shift the index of the sum to start with $l=0$.

• In (4) we apply the Chu-Vandermonde Identity.