Pascal’s triangle and cube numbers

To help explain where cubic numbers can be found in Pascal’s triangle, I will first briefly explain how square numbers are formed. The third diagonal of Pascal’s triangle is 1,3,6,10,15,21 … If we add 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 that cubic numbers can be formed from Pascal’s triangle is similar, but a little more complex. While square numbers can be found on the third diagonal, for cube numbers, we must look at the fourth diagonal. The first rows of Pascal’s triangle are shown below, with these numbers in bold:

eleven

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 twenty 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, the differences of which give the triangle numbers 1,3,6,10,15,21 (the sums of whole numbers, for example, 21 = 1 + 2 + 3 + 4 + 5). However, if you try to add consecutive pairs in the sequence 1, 4, 10, 20, 35, 56, you will not get the numbers from the cube. 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, you get (n ^ 3 + 3n ^ 2 + 2n ) / 6. We are basically trying to do n ^ 3, so a good starting point is that we have a term ^ 3/6 here, so we probably need to add six tetrahedral numbers to make n ^ 3, not 2. Try to find the cubic numbers from this information. If you’re still stuck, 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

In fact, this pattern always continues. If you want to see why this is the case, try expanding 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 n-th, (n-1) th, and (n-2) th tetrahedral numbers, and you should end up with n ^ 3. Otherwise As I hope to be the case (and I don’t blame you), just enjoy this interesting result and try it out with your friends and family to find out if they can spot this hidden link between Pascal’s triangle and cubic numbers.