Perhaps I have succeeded

We can 'cheat' and just write this down in summation notation: \( \sum_{i=1}^n \sum_{j=i+1}^n j-i \). If you remember how to do summations, great! But there are more interesting ways of thinking about this problem

The written description for n=4 gave the start of a recurrence solution, where the if you add point n you add the lines of length 1,2,3,.... n-2, n-1. These are the triangular numbers. So, our sum is in some (very real) sense stacking triangles of increasing size, which imply some 3 sided triangular pyramid of the numbers. I now know this is called the tetrahedral numbers.

However, there's another way of thinking about it. The visual description of n=4 imply that instead, we should first group our segments by length, and then add them to our sum. After checking back in with the picture for n=4, we see that there are 3 of length 1, 2 of length 2, and 1 off length 3. Extrapolating back into summation notation, this would be \( \sum_{i=1}^n i*(n-i) \). But more interestingly, this 'looks' like we're stacking rectangles that slowly morph into squares, and then back into rectangles in the other direction. This counts the same thing as before, so it should be the tetrahedral number- but how?

Before, we thought of tetrahedrons as starting from the base. But if instead you think of them as starting from the edge, it turns out that these rectangles to squares to rectangles is exactly the image we need. ( This makes me think of this problem, which somehow highlights the edge arrangement of a tetrahedron for me).

Here's the nice gif to show building a tetrahedron from triangles or rectangles! We're sticking with n=4, with the one on the left being building up triangles from the base and the one on the right building up rectangles from the edge:

I later discovered it can also be done purely combinatorically, which I think provides the 'most satisfying' answer, although like most combinatorics feels a little magical. Take a line with n+1 points, and offset it by 1/2 so it fully overlaps the original line. Then, for a segment on our line with n points, its 'subsegments of length one' overlaps one of our vertices on our line with n+1 points. The left vertex of the segment on n points corresponds to the point 1/2 to the left on the line with n+1 points, and the right vertex of the segment on n points corresponds to the point 1/2 to the right on the line with n+1 point. The picture below shows one line segment with n=4, with features on the top line colored and matching and the corresponding vertices in the bottom line with the additional vertex.

You can also find this on the tetrahedron, but its harder- I tend to think of it as an extension to the 'stacking triangles' case. The last point chooses how many triangles to go 'down' the tetrahedron from the vertex, and then the first 2 points count the lengths, in this clever way. I will say making sure that 'that point in the tetrahedron really exists' is a pain, but I've convinced myself.

This last model of the problem gives the answer \( {n+1 \choose 3} \)

I love this problem, largely because it was the first 'aha!" I had in math, particularly when connecting disparate subjects.