Start with the sequence of positive integers:
1, 2, 3, 4, …
Now take partial sums, the nth term of the new series being the sum of the first n terms of the previous series. This gives us the triangular numbers, so called because they count the number of coins at each level of a triangular arrangement:
1, 3, 6, 10, …
If we repeat this again we get the tetrahedral numbers, the number of balls on each level of a tetrahedral stack of balls.
1, 4, 10, 20, …
We can repeat this process and general define Tn, d, the nth tetrahedral number of dimension d, recursively. We define Tn, 1 = n and for d > 1,
This is just a formalization of the discussion above.
It turns out there’s a simple expression for tetrahedral number of all dimensions:
Here’s a quick proof by induction. The theorem clearly holds when n = 1 or d = 1. Now assume it hold for all n < m and d < m.
The last line follows from the binomial addition identity
It also turns out that Tn, d is the number of ways to select d things from a set of n with replacement.