# Pascal’s Triangle and Cube Numbers

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 past 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 complicate. 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 ordern 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 ordern 1,4,10,20,35,56, you do not get the cube numbers. To see how to get this ordern, 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 nevertheless stuck, then look at the next use.

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 consequence and test it out on your friends and family to find out if they can identify this hidden link between Pascal’s triangle and cube numbers!