# Find A Formula For 1 2 1 4 1 8 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 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 to the number before it, we get 0+1=1, 1+3=4, 3. +6=9, 6+10=16… , which are square numbers. The method of making cubic numbers from Pascal’s triangle is similar, but slightly more complicated. While square numbers can be found on the third diagonal, for cube numbers, we have to 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 (sums of whole numbers like 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 cubic numbers. To see how to achieve this order, 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 here we have the term ^3/6, so we need to add that together. is Tetrahedral numbers to make n^3, not 2. Try to find the cube numbers from this information. If you are still stuck, then see the next paragraph.

List the tetrahedral numbers with the first two zeros: 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, continue forever. If you want to see why this is the case, then 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)th 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 your friends and family to find out if they can find this hidden link between Pascal’s triangle and cubic numbers!

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