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How to show that $$ \binom {n+1} 3 + \binom {n+2} 3 = \sum_{i=1}^{n} i^2$$ by a combinatorial proof?

It's not difficult to check this algebraically and show that the left hand side is $\frac{n(n+1)(n-1)+n(n+1)(n+2)}{6} = \frac{n(n+1)(2n+1)}{6} = \sum_{i=1}^n i^2$. However, I fail to see how this is true by a combinatorial argument. In particular, I can visualise the right-hand side as a pile of cubes with each layer having $i^2$ cubes. However, I am not able to relate it to the left hand side. Any help will be appreciated!

Y.T.
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  • Maybe you can cast the pile of squares algebraically as the set of positive integers $i,j,k \le n$ such that $i \le k$ and $j \le k$. Then if I'm not mistaken, a partition into the cases $i < j$ and $i \ge j$ should do the trick. (But I haven't checked the details so I'm not going to post as an answer yet.) – Daniel Schepler Oct 12 '23 at 15:23
  • Note that $i^2=\binom{i}{2}+\binom{i+1}{2}$ and use the hockey stick identity. This can be made into a combinatorial proof, too. – Alexander Burstein Oct 13 '23 at 02:50

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