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.
Best Answer
$$\displaystyle\sum_{i=0}^k\sum_{j=0}^l\binom{i+j}i=\sum_{i=0}^k\sum_{j=i}^{i+l}\binom{j}i=\sum_{i=0}^k\binom{i+l+1}{i+1} \ ^{[1]}$$
$$=\sum_{i=0}^k\binom{i+l+1}{l}=\sum_{i=l}^{k+l+1}\binom{i}{l}−1=\binom{k+l+2}{k+1}-1\ ^{[1]}$$
1. Hockey-Stick Identity