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$T(n)$ 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.

Martin Sleziak
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theideasmith
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3 Answers3

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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.

Zubin Mukerjee
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5

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.

André Nicolas
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  • Maybe it is also interesting to mention that this is a special case of Hockey-Stick identity, where the lower index is $2$. http://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-t-0n-binom-tk-binomn1k1 http://www.artofproblemsolving.com/wiki/index.php/Combinatorial_identity#Hockey-Stick_Identity http://math.stackexchange.com/questions/linked/833451 – Martin Sleziak Feb 06 '16 at 11:29
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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 $$\begin{align} G_n=\sum_{j=1}^n\sum_{i=1}^ji &=\sum_{j=1}^n\sum_{i=1}^j\binom i1\\ &=\sum_{j=1}^n\binom {j+1}2\color{lightgrey}{=\sum_{j=1}^n T_n}\\ &=\binom {n+2}3\qquad\blacksquare \end{align}$$

Hypergeometricx
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