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## Pascal’s Triangle and Cube Numbers

To help explain where cube numbers can be found in Pascal’s triangle, I’ll 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 cube numbers can be formed from Pascal’s triangle is similar, but a bit more complex. While square numbers might be in the third diagonal, for cubic numbers you have to look at the fourth diagonal. The first 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 **ten** 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 corresponds to the tetrahedral numbers, the differences of which give the triangular 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 don’t 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, you get (n^3 + 3n^2 + 2n)/6. Basically we’re trying to do n^3 so a good starting point is here we have a ^3/6 term so we’ll probably have to add *six* tetrahedral numbers to make n^3, not 2. Try to find the cube numbers from this information. If you are still stuck, look at the next paragraph.

List the tetrahedral numbers with two leading zeros: 0,0,1,4,10,20,35,56…

Next, add three consecutive numbers at once, 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 × 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 still continues. If you want to see why this is the case, try extending 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 you will (and I don’t blame you), just enjoy this interesting result and test it out on your friends and family to see if they can spot this hidden link. between Pascal’s triangle and cubic numbers!

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