Navigation Panel: Previous | Up | Forward | Graphical Version | PostScript version | U of T Math Network Home

# Tetrahedral and 4-Tetrahedral Numbers

Asked by Warren Kendrick on November 4, 1997:
The tetrahedronal numbers: 1, 4, 10, 20, 35, ... are derived by adding the triangular numbers: 1 + 3 + 6 + 10 + 15 + ... + (n)(n+1)/2. At one time I derived a formula for the sum through the mth term of the tetrahedronal numbers, and also the sum of the numbers for a 4-tetrahedron: 1, 5, 15, 35, 70, ..., which are of course the sum of the tetrahedronal numbers. (Tetrahedronal numbers are best described by placing larger and larger triangular layers of 3-spheres under the previous triangular layer of spheres, forming a tetrahedronal pyramid.) Can you provide these formulas? Thanks.
These numbers are all expressible as binomial coefficients. If we let T_2(n) denote the nth triangular number, T_3(n) the nth tetrahedral number, and so on, then
```                        /  n+k-1  \       (n+k-1)!
T  (n)  =  (           )  =  --------- .
k          \    k    /      (n-1)! k!

```

You can prove this by induction noting that each T_k(1) equals 1, that the formula is correct when k=2, and the difference T_k(n+1)-T_k(n) = T_(k-1)(n+1).

You could also prove this by a more "brute-force" approach, using the formulas for the sums of powers:

S_1(n) = 1 + 2 + 3 + ... + n = n(n+1)/2

S_2(n) = 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6

S_3(n) = 1^3 + 2^3 + 3^3 + ... + n^3 = n^2(n+1)^2/4

and so on. To arrive at these formulas you can again proceed by brute force (make the intelligent guess that S_k(n) should be a polynomial of degree n+1 and plug in enough values to solve for the coefficients), or else employ a trick like the following:

Write (m+1)^(k+1) = m^(k+1) + (k+1) m^k plus terms involving lower powers of m. When you add up the LHS as m ranges from 0 to n, you get S_(k+1)(n+1) = S_(k+1)(n) + (n+1)^(k+1). When you add up the RHS, you get S_(k+1)(n) + (k+1) S_k(n) plus terms involving S_(k-1)(n) down through S_1(n). This gives you

S_(k+1)(n) + (n+1)^(k+1) = S_(k+1)(n) + (k+1) S_k(n) + lower order terms
The terms S_(k+1)(n) cancel, and you are left with a formula that expresses S_k(n) in terms of S_(k-1)(n) down through S_1(n).

For example, here is how to use the above technique to arrive at the formula for S_3(n):

(m+1)^4 = m^4 + 4 m^3 + 6 m^2 + 4 m + 1

S_4(n) + (n+1)^4 = S_4(n) + 4 S_3(n) + 6 S_2(n) + 4 S_1(n) + 1(n+1)

4 S_3(n) = (n+1)^4 - 6 S_2(n) - 4 S_1(n) - n - 1

S_3(n) = (1/4)[ (n+1)^4 - 6 n (n+1)(2n+1)/6 - 4 n(n+1)/2 - n - 1]

and simplifying gives the above formula. (Note that the reason n+1 appears at the end of the second line is that we are adding up the first line as m ranges from 0 to n; that means the final constant term "1" is added up n+1 times.)

Now that you have formulas for S_k(n), it is easy to find formulas for the tetrahedral numbers. For example,

T_2(n) = SUM_m<=n m(m+1)/2
= (1/2) SUM_m<=n m^2 + (1/2) SUM_m<=n m
= (1/2) n(n+1)(2n+1)/6 + (1/2) n(n+1)/2 = n(n+1)(n+2)/6.
[ Submit Your Own Question ] [ Create a Discussion Topic ]

This part of the site maintained by (No Current Maintainers)
Last updated: April 19, 1999
Original Web Site Creator / Mathematical Content Developer: Philip Spencer
Current Network Coordinator and Contact Person: Joel Chan - mathnet@math.toronto.edu