Supposedly the following proves the sum of the first-$n$-squares formula given the sum of the first $n$ numbers formula, but I don't understand it.

Enter image description here

Peter Mortensen
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4 Answers4


The first row of the first triangle is $1 = 1^2$, the second row sums to $2 + 2 = 2^2$, the third row sums to $3 + 3 + 3 = 3^2$ and so on. That means that the sum of the numbers in the triangle is $1^2 + 2^2 + 3^2 + \dots + n^2$. Now, the second and third triangles are the same, so the left-hand side is $3(1^2 + 2^2 + \dots + n^2)$.

On the other hand, each of the numbers in the right-hand side triangle is $(2n+1)$, and there are $\frac{n(n+1)}{2}$ of them.

The picture shows why the two are equal, so we get $3(1^2 + 2^2 + \dots +n^2) = (2n+1)\frac{n(n+1)}{2}$, which becomes the formula $1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$.

Michael Biro
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    Don't have pen and paper in hand, but do you think using four tetrahedrons would give formula of sum of cubed numbers? – Taozi Nov 15 '16 at 06:37
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    @Taozi: Offhand, it seems to me that to get a sum of cubes you need a square pyramid, not a tetrahedron. And since it's less symmetric you're more limited in the manipulations you can do. But there are other visual proofs for sum of cubes, e.g. https://en.wikipedia.org/wiki/Squared_triangular_number#/media/File:Nicomachus_theorem_3D.svg. – Meni Rosenfeld Nov 15 '16 at 09:42
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    The second and third triangles are the same as the first upon rotation by 120 degrees. This is a way cool demonstration! – richard1941 Nov 15 '16 at 17:54
  • @Taozi With four tetrahedra, each with $1$ at one vertex and $n$ at all $n(n+1)/2$ points on the opposite face, you would have $\binom n3$ points each containing the sum $3n+1$, so $\binom n3 (3n+1)=4\sum_{k=1}^n \frac12(k^3+k^2)=2\sum_{k=1}^n k^3+2\sum_{k=1}^n k^2$. Since we know $\sum_{k=1}^n k^2$ we could get $\sum_{k=1}^n k^3$ from this. Alternatively, in addition to the four tetrahedra labeled $1$ to $n$ along each edge, add four more tetrahedra labeled $2$ to $n$ along each edge so that the numbers in all tetrahedra add to $\sum_{k=1}^n k^3$. – David K Nov 27 '16 at 21:52

That first triangle is the sum of one $1$, two $2$s, three $3$s, and so on. That's equivalent to $1^2+2^2+\dots +n^2$.

The second triangle is just a $120^\circ$ counterclockwise rotation of the first triangle.

The third triangle is a $120^\circ$ clockwise rotation of the first triangle.

Now, note that the top numbers sum to $2n+1$. And the numbers directly below and to the left of each of the top numbers. And pretty much every single group of three numbers corresponding to the same position on the triangles.

So the RHS is just one $2n+1$, then two $2n+1$s, then three, and so on up to $n$. So the RHS is really $\frac{n(n+1)}{2}(2n+1)$ because of the triangular number formula.

This means that $3(1^2+2^2+\dots +n^2)=\frac{n(n+1)(2n+1)}{2}$, so $\boxed{1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}}.$

Edit: Beaten to it :P

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Each entry in the final triangle is the sum of the corresponding entries in the first three triangles. Each triangle has


entries, so the sum of all of the entries in the final triangle is


The entries in each of the first three triangles sum to

$$1\cdot1+2\cdot2+3\cdot3+\ldots+n\cdot n=\sum_{k=1}^nk^2\;,\tag{1}$$

where each term on the lefthand side of $(1)$ is the sum of the entries in one row of the triangle. Thus, the whole picture says that


and dividing through by $3$ yields the usual summation formula.

Brian M. Scott
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I don't need any votes for my answer, but for a different way to look at it, you can check http://www.maa.org/sites/default/files/Siu15722.pdf

Kirthi Raman
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