Pascals Triangle and Cube Numbers

Oct 9
08:48

2012

Alec Shute

An amazingly wide range of sequences can be found in Pascal's Triangle. In addition to triangle numbers, tetrahedral numbers, fibonacci numbers and square numbers, cube numbers a...

An amazingly wide range of sequences can be found in Pascal's triangle. In addition to triangle numbers, tetrahedral numbers, fibonacci numbers and square numbers, cube numbers are one sequence which is rarely discussed but can also be found in Pascal's triangle. In this article I explain where this sequence can be found in Pascal's triangle, and why these seemingly unrelated ideas are linked.

To help explain where cube numbers can be found in Pascal's triangle, I will first briefly explain how the square numbers are formed. The third diagonal in of Pascal's triangle is 1,3,6,10,15,21... If we add together each of these numbers with its previous number, we get 0+1=1, 1+3=4, 3+6=9, 6+10=16... , which are the square numbers. The way cube numbers can be formed from Pascal's triangle is similar, but a little more complex. Whilst the square numbers could be found in the third diagonal in, for the cube numbers, we must look at the fourth diagonal. The first few rows of Pascal's triangle are shown below, with these numbers in bold:

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

This sequence is the tetrahedral numbers, whose differences give the triangle numbers 1,3,6,10,15,21 (the sums of whole numbers e.g. 21 = 1+2+3+4+5). However, if you try adding up consecutive pairs in the sequence 1,4,10,20,35,56, you do not get the cube numbers. To see how to get this sequence, we will have to look at the formula for tetrahedral numbers, which is (n)(n+1)(n+2)/6. If you expand this, it you get (n^3 + 3n^2 + 2n)/6. Basically, we are trying to make n^3, so a good starting point is that here we have a n^3/6 term, so we are likely to need to add together six tetrahedral numbers to make n^3, not 2. Have a go at trying to find the cube numbers from this information. If you're still stuck, then look at the next paragraph.

List the tetrahedral numbers with two zeros first: 0,0,1,4,10,20,35,56...
Then, add three consecutive numbers at a time, but multiply the middle one by 4:

0 + 0 x 4 + 1 = 1 = 1^3

0 + 1 x 4 + 4 = 8 = 2^3

1 + 4 x 4 + 10 = 27 = 3^3

4 + 10 x 4 + 20 = 64 = 4^3

10 + 20 x 4 + 35 = 125 = 5^3

This pattern does in fact, always continue. If you want to see why this is the case, then try exanding and simplifying (n(n+1)(n+2))/6 + 4(n-1)(n)(n+1)/6 + ((n-2)(n-1)n)/6, which are the formulas for the nth, (n-1)th and (n-2)th tetrahedral numbers, and you should end up with n^3. Otherwise, as I expect is the case (and I don't blame you), just enjoy the this interesting result and test it out on your friends and family to find out if they can spot this hidden link between Pascal's triangle and cube numbers!

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