This series is about beautiful mathematics that needs only elementary tools.It is intended for anyone who likes math, but also for people who think they don’t like it. I take everything very slowly.

Today, I explore a beautiful pattern in adding consecutive integers.

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!

When 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. But induction confuses some people, 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 (n+1)² = n*n + n*1 + 1*n + 1*1 = n² + 2n + 1 and then we subtract 1 and get 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. Get a little frustrated but don’t drive yourself crazy.(A tolerance for frustration is key to math and to much else).

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 n²so that is (n — 1) n²s. So, we can take (n — 1) n²sfrom 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 or two on proofs

I 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.

Another thing that some may say is that, if I already knew one proof, why come up with another? Well, that’s like saying “if I already have one painting featuring a hillside, why paint another?” Proving things is a big part what mathematicians do. And just as different paintings have different things to say about a hillside, so different proofs have different things to say about math.

Finally, some may wonder why we need proofs at all. Isn’t the pattern “evident”? Well, things that are evident can be false. Once we prove something, we know it is true.



Besides, it’s fun. Play!

======

For another in this series see https://medium.com/peter-flom-the-blog/elementary-math-beautiful-math-the-highest-of-two-dice-d24c3afbd659

I will be tagging this series with the tag EMBM, so, if you want to follow along, you can.