# How To Do Citations In A Research Paper Apa Style 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 its previous number, we get 0+1=1, 1+3=4, 3 +6= 9, 6+10=16… , which are square numbers. The way in which 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 cubic numbers we must 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 10 5 1

1615 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, 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 in the cube. To see how to obtain 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 that here we have a term^3/6, so we’re likely going to have to add. six tetrahedral numbers to make n^3, not 2. Try finding the cubic numbers from this information. If you are still stuck, see 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 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 x 4 + 10 = 27 = 3^3

4 + 10 x 4 + 20 = 64 = 4^3

10 + 20 x 4 + 35 = 125 = 5^3

This pattern, in fact, always continues. If you want to see why this is so, 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 tetrahedral numbers, (n-1) and (n-2) and you should end up with n^3. If not, as I hope is the case (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 cubic numbers.

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