I think it's useful to report here another proof that I have posted on Mathoverflow.

Write down numbers in an equilateral triangle as follows:

    1
   2 2    
  3 3 3
 4 4 4 4

Now, clearly the sum of the numbers in the triangle is $Q_n:=1^2+2^2+\dots+n^2$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, like these ones

    1          4          4 
   2 2        3 4        4 3
  3 3 3      2 3 4      4 3 2
 4 4 4 4    1 2 3 4    4 3 2 1

then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$

The proof is not mine and I do not claim otherwise. I first heard it from János Pataki. It is similar (but simpler) to the proof appearing on Wikipedia as I am writing this.

How to prove formally that all positions sum to $2n+1$? Easy induction: moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases. This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well.

I think it's useful to report here another proof that I have posted on Mathoverflow.

Write down numbers in an equilateral triangle as follows:

    1
   2 2    
  3 3 3
 4 4 4 4

Now, clearly the sum of the numbers in the triangle is $Q_n:=1^2+2^2+\dots+n^2$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, like these ones

    1          4          4 
   2 2        3 4        4 3
  3 3 3      2 3 4      4 3 2
 4 4 4 4    1 2 3 4    4 3 2 1

then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$

The proof is not mine and I do not claim otherwise. I first heard it from János Pataki. It is similar (but simpler) to the proof appearing on Wikipedia as I am writing this.

How to prove formally that all positions sum to $2n+1$? Easy induction: moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases. This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well.

I think it's useful to report here another proof that I have posted on Mathoverflow.

Write down numbers in an equilateral triangle as follows:

    1
   2 2    
  3 3 3
 4 4 4 4

Now, clearly the sum of the numbers in the triangle is $Q_n:=1^2+2^2+\dots+n^2$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, like these ones

    1          4          4 
   2 2        3 4        4 3
  3 3 3      2 3 4      4 3 3
 4 4 4 4    1 2 3 4    4 3 2 1

then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$

The proof is not mine and I do not claim otherwise. I first heard it from János Pataki. It is similar (but simpler) to the proof appearing on Wikipedia as I am writing this.

How to prove formally that all positions sum to $2n+1$? Easy induction: moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases. This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well.

I think it's useful to report here another proof that I have posted on Mathoverflow.

Write down numbers in an equilateral triangle as follows:

    1
   2 2    
  3 3 3
 4 4 4 4

Now, clearly the sum of the numbers in the triangle is $Q_n:=1^2+2^2+\dots+n^2$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, like these ones

    1          4          4 
   2 2        3 4        4 3
  3 3 3      2 3 4      4 3 2
 4 4 4 4    1 2 3 4    4 3 2 1

then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$

The proof is not mine and I do not claim otherwise. I first heard it from János Pataki. It is similar (but simpler) to the proof appearing on Wikipedia as I am writing this.

How to prove formally that all positions sum to $2n+1$? Easy induction: moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases. This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well.