For an introduction to this series see this article. Briefly, I show beautiful mathematics that needs only elementary tools. In this article I showed a way to add consecutive integers, which is about as basic an operation as there is. You might think we are done with adding consecutive integers, but math has more beauty for us, even in this elementary realm.
Look at this:
- 1 + 2 = 3
- 4 + 5 + 6 = 7 + 8
- 9 + 10 + 11 + 12 = 13 + 14 + 15.
How amazing! How beautiful! What a pattern!
But, as I’ve said, once mathematicians see a pattern they like to prove that it continues. How could we prove this? If you like visual proofs, there is a nice one on page 3 of Charming Proofs by Claudi Alsina and Roger B. Nelson. They note that it can also be “proved by induction” but they don’t show such a proof. So here is a proof, taking things step by step (and small steps!)
Identifying the pattern
The first step is to figure out exactly what the pattern is. Let’s see. The first number in each line is 1, 4, 9… Seems like they are all square numbers. Let’s assume that for the moment (we will see proof of it along the way). So we have each line starting with n2. On the left side of the = sign we have 2 numbers in the first row, 3 in the second, 4 in the third….. That’s n ! Cool! Now we have the left side:
- n2 + (n2 + 1) + …. + (n2 + n).
What about the right side of the = sign? Well, it starts where the left side stopped, so that’s (n2 + n + 1). and each row has 1 fewer terms on the right than the left. Where does each right side end? Let’s see: Row 1 ends with 3, row 2 with 8, row 3 with 15. Each is (n+1)2 – 1, but a simpler way of writing that is n2 + 2n. The full system can be written:
- n2 + (n2 + 1) + … + (n2 + n) = (n2 + n + 1) + (n2 + n + 2) + ….+ (n2 + 2n)
How could we show these are equal? Try and think of ways! Play! Fool around before reading on.
Proving the pattern
As I said, Alsina and Nelson say this can be proved by induction, and it can. But here is another way.
How many n2 are on the left and the right?
On the left, we have n terms, and each has an n2 so that is n n2 s. On the right, we have n – 1 terms, again each has an n2 so that is (n – 1) n2 s. So, we can take (n – 1) n2s from each side leaving
- n2 + 1 + 2 + … + n = n + 1 + n + 2 + … + 2n
Next, rewrite 2n as (n + n):
- n2 + 1 + 2 + … + n = n + 1 + n + 2 + … + n + n.
How many n terms on the right? There are n + 1, but let’s just take n of them to get n times n, and n times n is….. n2 ! We can rearrange the terms to get:
- n2 + 1 + 2 + …. n = n2 + 1 + 2 + 3 + …. n
The two sides are the same!
A note on proofs
I just came up with that proof. One danger for students learning math is that they may think that proofs magically appear in the head of the mathematician, who simply writes them down. This is not how it works at all. Even for this relatively simple proof, I had to play around. I made errors. I went up wrong paths. I knew my destination was to get the two sides to look the same, and eventually I got there. More complex proofs may take hours, days or even years! But the proofs get written up in the books without the errors and wrong paths.
Play!
