For an introduction to this series, see this article. Briefly, I show some beautiful math that can be explained with only the math you learned in elementary school, giving lots of guidance for the math-phobic
In this article I looked at sums of consecutive integers and encouraged play. One way to modify that sum is to look at only odd numbers and see what happens.
The odd numbers, as you may know, are those in the series 1, 3, 5, …… But how can we characterize them without that annoying …. ? If we call the integers N and the odd integers D (Not using O because it is easily confused with 0) then every number in D is of the form 2N + 1. For example
- 1 = 2*0 + 1
- 3 = 2*1 + 1
- 5 = 2*2 + 1
- 101 = 2*50 + 1
Square numbers are those that are the squares of integers; that is, an integer multiplied by itself. They are written with a superscript 2. For example, 12 = 1*1 = 1. 22 = 2*2 = 4 and so on. They can also be represented by square arrays of dots; for example 22 is an array that is 2 dots on each side:
and there are four dots.
Adding the consecutive odd integers
Let’s start by adding some odd integers and see what happens:
- 1 = 1
- 1 + 3 = 4
- 1 + 3 + 5 = 9
- 1 + 3 + 5 + 7 = 16
Hmm. It seems like they are all square numbers! Cool! As I keep saying, mathematicians love patterns, but they also love to prove that the patterns continue. How could you prove that the series continues? Play for a while before reading on.
First proof that the sum of consecutive square integers is a square
Put a dot on a piece of paper (ideally, use graph paper, but regular paper will also work). That’s 1 dot and 1 is both an odd number and a square. What’s the next odd number? 3. So put 3 dots around the first dot. Put one to the right, one below and one diagonally below and to the right. Now you have 4 dots, arranged like the figure above. What’s the next odd number? 5. So put five dots around the four that you have, in the same pattern. Now you have 9 dots. And so on.
Second proof that the sum of consecutive odd integers is a square (this proof requires a little algebra)
We can look at the difference between n2 and (n+1)2. The second can be worked out as follows:
(n + 1)*(n+1) = (n+1)*n + (n+1)*1 = (n2+ n) + (n + 1) = n2 + 2n + 1. Subtracting n2 from this leaves 2n + 1 which is the next odd number (see above).
Play! What if you add even numbers? Or square numbers? Or something else? What if you don’t start at 1? Play!