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## Pascal’s Triangle and Square Numbers

Below are the first few rows of Pascal’s triangle:

1 1

1 2 **1**

1 3 **3** 1

1 4 **6** 4 1

1 5 **10** 10 5 1

1 6 **15** 21 15 6 1

The numbers in bold are the third diagonal in when Pascal’s triangle is drawn centrally. These are the triangle numbers, made from the sums of consecutive whole numbers (e.g. 15 = 1 + 2 + 3 + 4 + 5), and from these we can form the square numbers. All we have to do is add up consecutive numbers from these and we get the square numbers. To get the first square number, we have to add a 0 on to the front 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

Incidentally, you can also get the square numbers by taking the differences of numbers two places apart on the 4th diagonal in 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 the square numbers from adding together consecutive triangle numbers, you can use a variety of methods. Firstly, if you know that the formula for the nth triangle number is (n^2 + n)/2, then the previous triangle number is n less than this, as it’s the same sum of numbers but with (n-1) and not n as the last number you add. If we then add together 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 that method was not to your liking, we can also show this result pictorially. Triangle numbers derive their name from fact that you can make them by adding up the number of dots that make different sizes of triangle, and square numbers from the number of dots that make up different sized squares. So all we need to do is make a square from two triangles of dots. If you try this out 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 has to be one counter smaller on each of its sides. Okay, that wasn’t that rigorous a method for proving it, but it was a lot easier than doing a lot of algebra, wasn’t it?

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