Proof for this binomial coefficient's equation

Question

For $k, l \in \mathbb N$ $$\sum_{i=0}^k\sum_{j=0}^l\binom{i+j}i=\binom{k+l+2}{k+1}-1$$ How can I prove this?

I thought some ideas with Pascal's triangle, counting paths on the grid and simple deformation of the formula.

It can be checked here (wolframalpha).

If the proof is difficult, please let me know the main idea.

Sorry for my poor English.

Thank you.

EDIT: I got the great and short proof using Hockey-stick identity by Anubhab Ghosal, but because of this form, I could also get the Robert Z's specialized answer. Then I don't think it is fully duplicate.

Answer 1

$$\displaystyle\sum_{i=0}^k\sum_{j=0}^l\binom{i+j}i=\sum_{i=0}^k\sum_{j=i}^{i+l}\binom{j}i=\sum_{i=0}^k\binom{i+l+1}{i+1} \ ^{[1]}$$

$$=\sum_{i=0}^k\binom{i+l+1}{l}=\sum_{i=l}^{k+l+1}\binom{i}{l}−1=\binom{k+l+2}{k+1}-1\ ^{[1]}$$

1. Hockey-Stick Identity

Answer 2

Your idea about a combinatorial proof which is related to counting paths in a grid is a good one!

The binomial $\binom{i+j}i$ counts the paths on the grid from $(0,0)$ to $(i,j)$ moving only right or up. So the double sum $$\sum_{i=0}^k\sum_{j=0}^l\binom{i+j}i-1$$ counts the number of all such paths from $(0,0)$ to any vertex inside the rectangle $(0,k)\times (0,l)$ different from $(0,0)$.

Now consider the paths from $(0,0)$ to $(k+1,l+1)$ different from $$(0,0)\to (0,l+1)\to (k+1,l+1)\quad\text{and}\quad (0,0)\to (k+1,0)\to (k+1,l+1)$$ which are $$\binom{k+l+2}{k+1}-2.$$

Now any path of the first kind can be completed to a path of the second kind by changing direction, going to the boundary of the rectangle $(0,k+1)\times(0,l+1)$ and then moving to the corner $(k+1,l+1)$ along the side.

Is this a bijection between the first set of paths and the second one?

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{i = 0}^{k}\sum_{j = 0}^{\ell} {i + j \choose i} = {k + \ell + 2 \choose k + 1} - 1:\ {\LARGE ?}.\qquad k, \ell \in \mathbb{N}}$.

\begin{align} &\bbox[10px,#ffd]{\sum_{i = 0}^{k}\sum_{j = 0}^{\ell} {i + j \choose i}} = \sum_{i = 0}^{k}\sum_{j = 0}^{\ell}{i + j \choose j} = \sum_{i = 0}^{k}\sum_{j = 0}^{\ell}{-i - 1 \choose j} \pars{-1}^{\,j} \\[5mm] = &\ \sum_{i = 0}^{k}\sum_{j = 0}^{\ell}\pars{-1}^{\,j} \bracks{z^{\, j}}\pars{1 + z}^{-i - 1} = \sum_{i = 0}^{k}\sum_{j = 0}^{\ell}\pars{-1}^{\,j} \bracks{z^{0}}{1 \over z^{\, j}}\,\pars{1 + z}^{-i - 1} \\[5mm] = &\ \bracks{z^{0}}\sum_{i = 0}^{k}\pars{1 \over 1 + z}^{i + 1} \sum_{j = 0}^{\ell}\pars{-\,{1 \over z}}^{\,j} \\[5mm] = &\ \bracks{z^{0}}\braces{{1 \over 1 + z}\, {\bracks{1/\pars{1 + z}}^{k + 1} - 1 \over 1/\pars{1 + z} - 1}} \braces{{\pars{-1/z}^{\ell + 1} - 1 \over -1/z - 1}} \\[5mm] = &\ \bracks{z^{0}}\braces{% {1 - \pars{1 + z}^{k + 1} \over -z} \,{1 \over \pars{1 + z}^{k + 1}}} \braces{{\pars{-1}^{\ell + 1} - z^{\ell + 1} \over -1 - z}\,{z \over z^{\ell + 1}}} \\[5mm] = &\ \bracks{z^{\ell + 1}}\braces{1 - {1 \over \pars{1 + z}^{k + 1}}} \braces{z^{\ell + 1} + \pars{-1}^{\ell} \over 1 + z} \\[5mm] = &\ \pars{-1}^{\ell}\bracks{z^{\ell + 1}} \bracks{\pars{1 + z}^{-1} - \pars{1 + z}^{-k - 2}} \\[5mm] = &\ \pars{-1}^{\ell}\bracks{\pars{-1}^{\ell + 1} - {-k - 2 \choose \ell + 1}} \\[5mm] = &\ -1 - \pars{-1}^{\ell}\,{-\bracks{-k - 2} + \bracks{\ell + 1} - 1 \choose \ell + 1}\pars{-1}^{\ell + 1} \\[5mm] = &\ -1 + { k + \ell + 2 \choose \ell + 1} = \bbx{{k + \ell + 2 \choose k + 1} - 1} \end{align}