<s>[INST] 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. 
 [/INST] If one is in a combinatorial mood, one can do it basically without computation, as follows.
Note that $\frac{(n+1)(n)}{2}=\binom{n+1}{2}$. Note also that $\frac{(x^2+x)(x+2)}{6}=\binom{x+2}{3}$.
So we want to show that
$$\sum_1^x \binom{n+1}{2}=\binom{x+2}{3}.\tag{1}$$
But Formula 1 has a simple combinatorial interpretation. Imagine that we are choosing $3$ numbers from the numbers $1,2,\dots,x+2$. There are clearly $\binom{x+2}{3}$ ways to do it. That gives us the right-hand side of Formula 1. Now  let us count the number of choices another way.
There are $\binom{x+1}{2}$ choices where $1$ is the smallest number chosen. For the other two numbers can be chosen from the remaining $x+1$ in $\binom{x+1}{2}$ ways.
There are $\binom{x}{2}$ choices where $2$ is the smallest number chosen. For the other two numbers can be chosen from the remaining $x$ in $\binom{x}{2}$ ways.
And so on. Finally, there are $\binom{2}{2}$ ways to choose so that $x$ is the smallest number chosen.
Add up. We get the left-hand side of Formula 1.
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