Tetrahedral numbers

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 T_{n, d}, the nth tetrahedral number of dimension d, recursively. We define T_{n, 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 T_{n, d} is the number of ways to select d things from a set of n with replacement.

Related posts:

• When is a triangle a square?
• Sums of consecutive powers