All The Formulas You Need To Know For Math Ii Pascal’s Triangle and Square Numbers

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Pascal’s Triangle and Square Numbers

Below are the first rows of Pascal’s triangle:

1 1

1 2 1

1 3 3 1

1 4 6 4 1

15 10 10 5 1

16 15 21 15 6 1

The numbers in bold are the third diagonal when Pascal’s triangle is drawn in the center. These are the triangular numbers, made from the sums of consecutive integers (eg 15 = 1 + 2 + 3 + 4 + 5), and from these we can form the square numbers. All we have to do is add consecutive numbers of these and we get the square numbers. To get the first square number, we need to add a 0 to the top of the list:

0, 1, 3, 6, 10, 15, 21…

0 + 1 = 1 = 1^2

1 + 3 = 4 = 2^2

3+ 6 = 9 = 3^2

6 + 10 = 16 = 4^2

10 + 15 = 25 = 5^2

15 +21 = 36 = 6^2

By the way, you can also get the squared numbers by taking the differences of numbers at two separate places on the fourth diagonal of Pascal’s triangle. The fourth diagonal goes 1, 4, 10, 20 35… , and the differences you get are 1-0 = 1, 4-0 = 4, 10-1 = 9, 20-4 = 16, 35-10 = 25 and so on.

To understand why you get square numbers by adding consecutive triangular numbers, you can use several methods. First, if you know that the formula for the nth number in the triangle is (n^2 + n)/2, then the number in the previous triangle is n less than this one, since it is the same sum of numbers but with ( n-1) and not n as the last number you add. If we add these two numbers, we get

(n^2 + n)/2 + (n^2 + n)/2 – n

= (1/2)n^2 + n/2 + (1/2)n^2 + n/2 – n

= n^2 + n – n

= n^2

If this method was not to your liking, we can also show this result pictorially. Triangle numbers get their name from the fact that you can make them by adding up the number of points that make different sized triangles, and square numbers from the number of points that make squares of different sizes. So all we have to do is make a square from two point triangles. If you try this with coins or counters, or on paper, and make right-angled triangles, you should find that you can make a square from two triangles, but one must be a smaller counter on each of its sides. Okay, so it wasn’t such a rigorous method of proving it, but it was a lot easier than doing a lot of algebra, right?

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