# 2 4 Is Same As 2 X 4 In Math Pascal’s Triangle and Cube Numbers

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## Pascal’s Triangle and Cube Numbers

To help explain where cubic 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 to its previous number, we get 0+1=1, 1+3=4, 3 + 6=9, 6+10=16… , which are the square numbers. The way cubic numbers can be formed from Pascal’s triangle is similar, but slightly more complex. While square numbers can be found on the third diagonal, for cubic numbers, we 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

1510 10 5 1

1 6 15 20 15 6 1

1721 35 35 21 7 1

1828 56 70 56 28 8 1

This sequence is 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 will not get the numbers in the cube. To see how to get this sequence, we need to look at the formula for tetrahedral numbers, which is (n)(n+1)(n+2)/6. If you expand it, you get (n^3 + 3n^2 + 2n)/6. Basically, we’re trying to make n^3, so a good starting point is that we have a ^^3/6 term here, so we’re likely going to have to add together. six tetrahedral numbers to make n^3, not 2. Try to find the cubic numbers from this information. If you are still stuck, see the next paragraph.

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

Then add three consecutive numbers together, 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 continues forever. If you want to see why this is, 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 nth, (n-1) and (n-2)th tetrahedral numbers, and you should end up with n^3. Otherwise, as I expected (and I don’t blame you), just enjoy this interesting result and test it with your friends and family to see if they can spot this hidden link between Pascal’s triangle and the numbers in the cube !

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