Harold Cooper
8th August 2022
This article discusses a pattern I noticed in ‘visual’ derivations of the formulas for and , which led me to a similar derivation for .
using two lines
There’s a well-known trick for adding . Leaving out a bunch of symbols, it looks like this:
In other words, we arrange the sum as a line, then add a flipped copy of the line, multiplying by to keep the total the same.
Since consecutive entries increase by in the first line but decrease by in the flipped line, the combined entries all have the same value.
So we have entries all of value , and can simply multiply to get the total.
using three triangles
Triangle arrangement of
I recently encountered1 a similar trick for the sum of squares , but this time using three triangles instead of two lines!
In other words, we arrange the sum as a triangle—one one (), followed by two twos (), and so on, to the last row of ’s ().
We then add two rotated copies of the triangle so we have all three orientations (i.e. the gets to be at each of the three corners), multiplying by to keep the total the same.
Every entry in the resulting triangle has the same value of 2, and there are (as derived above!) entries, so we simply multiply to get the total.
using four tetrahedra
Since this trick worked for using lines and using triangles, I wanted to see if any shape would work for the sum of cubes .
Pyramids?
The simplest way to arrange is as a pyramid, where the top layer is one one (), the second layer is two-by-two twos (), and so on, to the last layer of -by- ’s (). For example, for :
Pyramid arrangement of
But pyramids aren’t very symmetrical—the sides are triangles but the base is a square, so every symmetry leaves the at the top and doesn’t actually change the entries at all, meaning we can’t combine copies in a helpful way.
Octahedra?
If you double the pyramid, you get a much more symmetrical object—the octahedron. It represents (two pyramids, minus one th layer since it isn’t doubled). For example, for :
Octahedron arrangement of
This looks promising, since we can combine rotated copies of it as we did with lines and triangles. But this only helps if the combined entries all have the same value, and it turns out they don’t. For , for example:
Combining the three unique rotations of the octahedron
The combined entries aren’t all equal—for example, the top is but the center is .
Tetrahedra!
Doubling the pyramid didn’t work, so maybe we can cut it in half instead!
It turns out the entries of our pyramid are equivalent to the entries of two tetrahedra, minus the shared central triangle. We can visualize this for for example:
And the pattern holds for all :
We already know the formula for the triangle—it’s just from above!—and tetrahedra are much more symmetrical than pyramids, so we can try the rotated copies trick again:
In other words, we add three rotated copies of the tetrahedron so we have all four orientations (i.e. the gets to be at each of the four corners), multiplying by to keep the total the same.
Every entry in the resulting tetrahedron has the same value of 3, and there are entries4, so we simply multiply to get the total.
And lastly we just substitute the formula for the triangle (derived above!) and simplify the polynomial.
Summary
So there we have it, all the ‘visual’ sums of powers before you need more than three dimensions, which isn’t very visual for humans.
In the context of simplices, going from 2 line segments to 3 triangles to 4 tetrahedra is a nice pattern—line segments are 1-simplices, triangles are 2-simplices, and tetrahedra are 3-simplices.
The pattern can continue, using 5 four-dimensional 4-simplices to derive the formula for , and so on in increasingly high dimensions. But that might defeat the point of it being a ‘visual’ derivation.
…but also
If we write the formula for in terms of the formula for , an interesting identity emerges, known as Nicomachus’s Theorem:
We can express this visually, as in this image by Wikipedia user cmglee:
Visual proof that [source]
This is a much easier way to visually derive the formula for , but don’t worry I still had fun figuring out the tetrahedron way—the more, the merrier!
Addendum If you’re curious how I made the spinning diagrams in this article, I’ve since written an explanation.