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. 
 A: 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. 
A: 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.
A: Natural numbers:
$1, \quad 2,\quad 3, \quad4, \cdots $
Triangular numbers:
$1,\quad 1+2,\quad 1+2+3,\quad  1+2+3+4,\quad \cdots $
Tetrahedral numbers:
$1,\quad 1+(1+2),\quad  1+(1+2)+(1+2+3),\quad  1+(1+2)+(1+2+3)+(1+2+3+4),\cdots$ 
Hence the $n$-the Tetrahedral number, $G_n$ is given by
$$\begin{align}
G_n=\sum_{j=1}^n\sum_{i=1}^ji
&=\sum_{j=1}^n\sum_{i=1}^j\binom i1\\
&=\sum_{j=1}^n\binom {j+1}2\color{lightgrey}{=\sum_{j=1}^n T_n}\\
&=\binom {n+2}3\qquad\blacksquare
\end{align}$$
