summation - Deriving the formula for the $n^{th}$ tetrahedral number

$T(n)$ is $n^{th}$ triangular number, where $T\left(n\right)=\frac{n^2+n}{2}$

And from other sources I know the $n^{th}$ tetrahedral number is

$G\left(x\right)=\sum _{n=1}^xT\left(n\right)=\frac{\left(x^2+x\right)\left(x+2\right)}{6}$

I also happen to know that the formula for the volume of a tetrahedron is:

$V=\frac{1}{2}Ah$,

where $A$ is the area of the triangular base and $h$ is the height.

If I sat down one day not knowing the formula for $G(x)$ and wanted to create a function to find the $n^{th}$ tetrahedral number, how do I derive it?

I've seen proofs. I want to know how the proof authors arrived at that formula in the first place.

Answer

In general, if $f(n)$ is a polynomial with degree $k$, and if $$\sum_{x=1}^n f(x) = g(n)$$

then $g(n)$ must be a polynomial with degree $k+1$.

---

This means that since triangular numbers are given by a polynomial of degree $2$, tetrahedral numbers must be given by a polynomial of degree $3$.

Let $a,b,c,d \in \mathbb{R}$ such that the $n^\text{th}$ tetrahedral number $G(n)$ is given by

$$an^3 + bn^2 + cn + d$$

We know immediately that $d=0$ because $G(0)=0$ (the empty sum).

Now we can simply list out any three tetrahedral numbers to find the general formula. Let's use the first three.

$$\begin{align*} G(1) \,&=\, T(1) = 1 \\\\ G(2) \,&=\,T(1) + T(2) = 1 + 3 \\\\ G(3) \,&=\,T(1) + T(2) + T(3) = 1 + 3 + 6 \\ \end{align*}$$

Rewriting, we get these equations involving the coefficients:

$$\begin{align*} 1 &= a + b + c\\\\ 4 &= 8a + 4b + 2c\\\\ 10 &= 27a + 9b + 3c\\ \end{align*}$$

Three linear equations, three variables. While it's a bit tedious to do the row reduction, it's definitely one way to derive the formula.