# A combinatorial identity - Hockey Stick generalization

There is a well known identity (the so called "Hockey-stick identity") asserting that:

$$\sum_{j=0}^m \binom{r+j}{j} = \binom{m+r+1}{r+1}$$

For some proofs see this.

I need to prove a kind of generalization, namely:

$$\sum_{j=0}^m \binom{r+j}{j}\binom{s+j}{j} = \sum_{j=0}^s \binom{r}{j}\binom{s}{j} \binom{m+r+s+1-j}{r+s+1}$$ For every $$r\geq s\geq 0$$.

Of course, setting $$s=0$$ in the latter gives the original identity. The problem is that I'm not being able to prove the second using the same ideas as those that work for the first one.

Any kind of help would be very appreciated.

## 3 Answers

\begin{align} \sum_{j=0}^m\binom{r+j}{j}\binom{s+j}{j} &=\sum_{j=0}^m\sum_{k=0}^r\binom{r}{k}\color{#C00}{\binom{j}{k}\binom{s+j}{j}}\tag1\\ &=\sum_{j=0}^m\sum_{k=0}^r\binom{r}{k}\color{#C00}{\binom{s+k}{k}\binom{s+j}{s+k}}\tag2\\ &=\sum_{k=0}^r\binom{r}{k}\binom{s+k}{k}\binom{s+m+1}{s+k+1}\tag3\\ &=\sum_{k=0}^r\sum_{j=0}^s\color{#C00}{\binom{r}{k}}\binom{s}{j}\color{#C00}{\binom{k}{j}}\binom{s+m+1}{s+k+1}\tag4\\ &=\sum_{k=0}^r\sum_{j=0}^s\color{#C00}{\binom{r}{j}}\binom{s}{j}\color{#C00}{\binom{r-j}{k-j}}\binom{s+m+1}{s+k+1}\tag5\\ &=\sum_{j=0}^s\binom{r}{j}\binom{s}{j}\binom{m+r+s+1-j}{r+s+1}\tag6\\ \end{align} Explanation:
$$(1)$$: Vandermonde's Identity: $$\binom{r+j}{j}=\sum_k\binom{r}{k}\binom{j}{j-k}$$
$$(2)$$: expand red binomial coefficients as ratios of factorials
$$(3)$$: sum in $$j$$ using the Hockey-Stick Identity
$$(4)$$: Vandermonde's Identity: $$\binom{s+k}{k}=\sum_j\binom{s}{j}\binom{k}{k-j}$$
$$(5)$$: expand red binomial coefficients as ratios of factorials
$$(6)$$: $$\binom{r-j}{k-j}=\binom{r-j}{r-k}$$, then Vandermonde's Identity

• Clever approach. (+1) Commented Feb 3, 2020 at 7:40

With the convention $$\binom{x}{n}=x^{\,\underline {\,n\,} } /n!$$, in the RHS $$r+s+1$$ shall be a non-negative integer.
We can then apply symmetry of the binomial to rewrite our identity as $$\sum\limits_{j = 0}^m {\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)} = \sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)} = \sum\limits_{j = 0}^s {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ m + r + s + 1 - j \cr r + s + 1 \cr} \right)} = \sum\limits_j^{} {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ m + r + s + 1 - j \cr m - j \cr} \right)}$$ i.e.: $$\bbox[lightyellow] { \sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)} = \sum\limits_j^{} {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ m + r + s + 1 - j \cr m - j \cr} \right)} \tag{1} }$$

A possible way to demonstrate it is by taking the ogf over $$m$$ as follows

For the LHS \eqalign{ & G_{\,a} (z) = \sum\limits_{0\, \le \,m} {\sum\limits_{j = 0}^m {\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;m} } } = \cr & = \sum\limits_{0\, \le \,m} {\sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;m} } } = \cr & = \sum\limits_{0\, \le \,m} {\sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)z^{\;m - j} \left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;j} } } = \cr & = \left( {\sum\limits_{0\, \le \,m} {\left( \matrix{ m \cr m \cr} \right)z^{\;m} } } \right)\left( {\sum\limits_{0\, \le \,m} {\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;j} } } \right) \cr & = {1 \over {1 - z}}\sum\limits_{0\, \le \,k} {\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)z^{\;j} } = \cr & = {1 \over {1 - z}}\sum\limits_{0\, \le \,k} {{{\left( {r + 1} \right)^{\,\overline {\,k\,} } \left( {s + 1} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k\,} } }}{{z^{\;k} } \over {k!}}} = \cr & = {1 \over {1 - z}}{}_2F_{\,1} \left( {\left. {\matrix{ {r + 1,s + 1} \cr 1 \cr } \;} \right|\;z} \right) \cr}

For the RHS \eqalign{ & G_{\,b} (r,s,z) = \sum\limits_{0\, \le \,m} {\sum\limits_j^{} {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)\left( \matrix{ r + s + 1 + m - j \cr m - j \cr} \right)z^{\;m} } } = \cr & = \sum\limits_{0\, \le \,m} {\sum\limits_j^{} {\left( \matrix{ r \cr j \cr} \right)\left( \matrix{ s \cr j \cr} \right)z^{\;j} \left( \matrix{ r + s + 1 + m - j \cr m - j \cr} \right)z^{\;m - j} } } = \cr & = \left( {\sum\limits_{0\, \le \,k} {\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)z^{\;k} } } \right)\sum\limits_{0\, \le \,k} {\left( \matrix{ r + s + 1 + k \cr k \cr} \right)z^{\;k} } = \cr & = \left( {\sum\limits_{0\, \le \,k} {\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)z^{\;k} } } \right){1 \over {\left( {1 - z} \right)^{\;r + s + 2} }} = \cr & = {1 \over {\left( {1 - z} \right)^{\;r + s + 2} }}\sum\limits_{0\, \le \,k} {\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)z^{\;k} } = \cr & = {1 \over {\left( {1 - z} \right)^{\;r + s + 2} }}\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\;k} \left( { - 1} \right)^{\;k} \left( \matrix{ k - r - 1 \cr k \cr} \right)\left( \matrix{ k - s - 1 \cr k \cr} \right)z^{\;k} = } \cr & = {1 \over {\left( {1 - z} \right)^{\;r + s + 2} }}\sum\limits_{0\, \le \,k} {{{\left( { - r} \right)^{\,\overline {\,k\,} } \left( { - s} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k\,} } }}{{z^{\;k} } \over {k!}}} = \cr & = {1 \over {\left( {1 - z} \right)^{\;r + s + 2} }}{}_2F_{\,1} \left( {\left. {\matrix{ { - r, - s} \cr 1 \cr } \;} \right|\;z} \right) \cr}

The Euler transformation for the Hypergeometric gives $${}_2F_{\,1} \left( {\left. {\matrix{ { - r, - s} \cr 1 \cr } \;} \right|\;z} \right) = \left( {1 - z} \right)^{\,1 + r + s} {}_2F_{\,1} \left( {\left. {\matrix{ {r + 1,s + 1} \cr 1 \cr } \;} \right|\;z} \right)$$ which completes the demonstration.

It is interesting to note that the sides in id. 1 are polynomials in $$r,s$$ of degree $$m,m$$.

Therefore the identity holds as well for real and even complex values of $$r, \,s$$.

Furthermore (I realized just now going through my notes on binomial identities) it can be straightly deduced from this other basic identity \bbox[lightyellow] { \eqalign{ & \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)\,} {( - 1)^{\,m - k} \left( \matrix{ x + y + 1 \cr m - k \cr} \right)\left( \matrix{ x + k \cr k \cr} \right)\left( \matrix{ y + k \cr k \cr} \right)} = \left( \matrix{ x \cr m \cr} \right)\left( \matrix{ y \cr m \cr} \right)\quad \Leftrightarrow \cr & \Leftrightarrow \quad \left( \matrix{ x + m \cr m \cr} \right)\left( \matrix{ y + m \cr m \cr} \right) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ x \cr k \cr} \right)\left( \matrix{ y \cr k \cr} \right)\left( \matrix{ x + y + m - k \cr m - k \cr} \right)} \quad \left| \matrix{ \;{\rm integer}\,m \hfill \cr \,x,y \in C \hfill \cr} \right. \cr} \tag{2} } that is called Suranyi's formula in this paper, and demonstrated therein.
It has also been dealt with and demonstrated in this related post.

In fact, from the above we get \eqalign{ & \sum\limits_j^{} {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r + j \cr j \cr} \right)\left( \matrix{ s + j \cr j \cr} \right)} = \cr & = \sum\limits_j^{} {\sum\limits_k {\left( \matrix{ m - j \cr m - j \cr} \right)\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)\left( \matrix{ r + s + j - k \cr j - k \cr} \right)} } = \cr & = \sum\limits_k^{} {\left( \matrix{ r \cr k \cr} \right)\left( \matrix{ s \cr k \cr} \right)\left( \matrix{ r + s + m + 1 - k \cr m - k \cr} \right)} \cr}

Using the symmetry $$\binom{p}{q}=\binom{p}{p-q}$$ of binomial coefficients, we want to show for non-negative integers $$m$$ and integers $$0\leq r\leq s$$: \begin{align*} \sum_{k=0}^m \binom{r+k}{r}\binom{s+k}{s} = \sum_{j=0}^s \binom{r}{j}\binom{s}{j} \binom{m+r+s-j+1}{r+s+1}\tag{1} \end{align*}

At first we look at the right-most binomial coefficient in (1) where we can apply the Hockey-stick identity. The summand $$+1$$ indicates a telescoping approach via $$\binom{p+1}{q+1}-\binom{p}{q+1}=\binom{p}{q}$$. Indeed, we obtain \begin{align*} \sum_{k=0}^m\binom{r+s-j+k}{r+s}&=\sum_{k=0}^m\left(\binom{r+s-j+k+1}{r+s+1}-\binom{r+s-j+k}{r+s+1}\right)\\ &=\binom{m+r+s-j+1}{r+s+1}\tag{2} \end{align*} where the telescoping sum permits cancellation of all terms besides the first and the last summand and the first summand vanishes also since it is $$\binom{r+s-j}{r+s+1}=0$$.

Using (2) we can now write the claim (1) in the form \begin{align*} \sum_{k=0}^m \binom{r+k}{r}\binom{s+k}{s} = \sum_{k=0}^m \sum_{j=0}^s \binom{r}{j}\binom{s}{j} \binom{m+r+s-j}{r+s} \tag{3} \end{align*}

In fact we can show that we have termwise equality in (3) for each $$0\leq k\leq m$$: \begin{align*} \binom{r+k}{r}\binom{s+k}{s} = \sum_{j=0}^s \binom{r}{j}\binom{s}{j} \binom{m+r+s-j}{r+s} \tag{4} \end{align*}

The identity (4) is named after Surányi (1955) which has already been mentioned by @GCab. We show the validity of (4) by closely following an approach given in Combinatorial identities by J. Riordan.

We start with the left-hand side of (4) and obtain \begin{align*} \color{blue}{\binom{r+k}{r}}&\color{blue}{\binom{s+k}{s}}\\ &=\sum_{j=0}^r\binom{r}{r-j}\binom{k}{j}\binom{s+k}{s}\tag{5}\\ &=\sum_{j=0}^r\binom{r}{r-j}\binom{s+j}{j}\binom{s+k}{s+j}\tag{6}\\ &=\sum_{j=0}^r\binom{r}{r-j}\binom{s+j}{j}\sum_{l=0}^{r-j}(-1)^{r-j-l}\binom{s+k+l}{s+r}\binom{r-j}{l}\tag{7}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\sum_{j=0}^{r-l}(-1)^{r-j-l}\binom{s+j}{j}\binom{r}{r-j}\binom{r-j}{l}\tag{8}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\binom{r}{l}\sum_{j=0}^{r-l}(-1)^{-j-l}\binom{s+j}{j}\binom{r-l}{r-l-j}\tag{9}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\binom{r}{l}(-1)^{r-l}\sum_{j=0}^{r-l}\binom{-s-1}{j}\binom{r-l}{r-l-j}\tag{10}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\binom{r}{l}(-1)^{r-l}\binom{-s-1+r-l}{r-l}\tag{11}\\ &=\sum_{l=0}^r\binom{s+k+l}{s+r}\binom{r}{l}\binom{s}{r-l}\tag{12}\\ &\,\,\color{blue}{=\sum_{l=0}^r\binom{r}{l}\binom{s}{l}\binom{r+s+k-l}{r+s}}\tag{13} \end{align*} and the claim (4) follows for $$0\leq k\leq m$$ and so also the claim (1).

Comment:

• In (5) we apply Vandermonde's identity to $$\binom{r+k}{r}$$.

• In (6) we use the binomial identity $$\binom{k}{j}\binom{s+k}{s}=\binom{s+j}{j}\binom{s+k}{s+j}$$.

• In (7) we use the binomial identity $$\binom{n}{m}=\sum_{l=0}^r(-1)^{r-l}\binom{n+l}{m+r}\binom{r}{l}$$ which is shown at the end of this post.

• In (8) we exchange the sums.

• In (9) we use the binomial identity $$\binom{r}{r-j}\binom{r-j}{l}=\binom{r}{l}\binom{r-l}{r-l-j}$$.

• In (10) we apply the binomial identity $$\binom{-p}{q}=(-1)^q\binom{p+q-1}{q}$$ to $$\binom{s+j}{j}$$.

• In (11) we apply again Vandermonde's identity.

• In (12) we use again $$\binom{-p}{q}=(-1)^q\binom{p+q-1}{q}$$.

• In (13) we change the order of summation $$l\to r-l$$.

Proof of (7): \begin{align*} \binom{n}{m}=\sum_{l=0}^r(-1)^{r-l}\binom{n+l}{m+r}\binom{r}{l}\tag{14} \end{align*}

We obtain \begin{align*} \color{blue}{\binom{n}{m}}&=(-1)^m\binom{-n+m-1}{m}\tag{12}\\ &=(-1)^m\sum_{l}\binom{-n+m-1-r}{l}\binom{r}{m-l}\tag{11}\\ &=\sum_{l}(-1)^{m+l}\binom{n-m+r+l}{l}\binom{r}{m-l}\tag{12}\\ &=\sum_{l}(-1)^l\binom{n+r-l}{m-l}\binom{r}{l}\tag{15}\\ &=\sum_{l}(-1)^l\binom{n+r-l}{n-m-l}\binom{r}{l}\tag{16}\\ &=\sum_{l}(-1)^l\binom{n+r-l}{m+r}\binom{r}{l}\tag{17}\\ &\,\,\color{blue}{=\sum_{l}(-1)^{r-l}\binom{n+l}{m+r}\binom{r}{l}}\tag{13} \end{align*} and the claim (14) follows.

Comment:

• In (15) we change the order of summation $$l\to m-l$$.

• In (16) we set $$m\to n-m$$, since $$\binom{n}{m}=\binom{n}{n-m}$$.

• In (17) we use $$\binom{p}{q}=\binom{p}{p-q}$$.