Odd thing I noticed about a simple mathematical pattern

I was carrying my sleepy cranky four-year-old up a hiking trail yesterday, which gave my brain time to wander, and I noticed something.

Take the common pattern made with adding things to a triangle, where you create a new row by increasing the number of things in the row by 1 each time, like so:


and so on.

As you go down the triangle, your running total looks like this:

and so on.

With me so far?

So I noticed that between the first and second row, your running total is multiplied by 3 (1 x 3 = 3). Between the second and third rows, your running total is multiplied by 2 (3 x 2 = 6). Between the third and fourth it got tricky.

But here’s what I noticed:

1 x 3/1 = 3
2 x 4/2 = 6
6 x 5/3 = 10
10 x 6/4 = 15
15 x 7/5 = 21
21 x 8/6 = 28
28 x 9/7 = 36

And, presumably, so on.

In other words, if you add one to the numerator and denominator of the previous fraction and multiply your current running total by the resulting number, you’ll get the next running total.

This blew my mind, because at first glance, the two patterns ({1, 3, 6, 10, 15} and {3/1, 4/2, 5/3, 6/4, 7/5}) are completely unrelated–and the second pattern isn’t one I’ve ever seen elsewhere. But surely there’s a reason for it.

I’m decent at understanding math, but my vocabulary is pretty limited. Can anyone explain in very small words why this pattern works?

All numerators except the last two cancel out with all denominators except the first 2, so you’ll end up with n * (n + 1) / 2.

I’m afraid I don’t follow. Do you mind using littler words, or making explicit the things you’re thinking are obvious?

The Gaussian sum of the first numbers from 1 through n is n * (n + 1) / 2.

Now if you derive the n-th term of the series by starting at 1 and multiplying all the ratios 3/1 * 4/2 * 5/3 * … * n/(n-2) * (n+1)/(n-1), you’ll find most numbers cancelling out and also leaving n * (n + 1) / 2.

The formula, in general, for the sum of the items in that triangle, is n(n+1)/2

So, for the sum of numbers 1 through, say, 7, you get 7*8 / 2 = 28

Now, bump up to the next level. The sum will be (n+1)*(n+2)/2.

And the ratio of those two numbers – (n+1)(n+2)/2 / n(n+1)/2 – is (n+2)/n

So if 28 is the 7th sum, 28 * 9/7 = 36, which is the eighth sum. 36 * 10/8 = 45 which is the ninth sum, and so on.

You can specify the items explicitly – n(n+1)/2 – or incrementally – N[n+1] = N[n] * n+2/n

A little algebra shows that each of these two ways to look at the series can be manipulated to reveal the other. Lovely stuff!

Gauss, it is said, figured this out by cutting the triangle in half and putting the two parts together as a rectangle.

Huh. I think I need to play with this a bit to understand it.

I thought the story was that he figured it out by folding the list of numbers being added back upon itself, and noting that the vertical sums were identical. At the age of seven, or something like that. :eek:

I think the simplest way to state this is:

The sum of 1 + 2 + …+ n = n(n+1)/2
The sum of 1 + 2 + …+ n + n+1 = (n+1)(n+1+1)/2
The ratio is clearly (n+2)/n which is what you found.

Knowing Gauß more likely he figured out half a dozen different proofs and wrote them down in his notebook.

Patterns of this sort occur with all the terms in the series for sums of powers (that is, in the formula for sums of the first n squares, first n cubes, etc.) I just learned this about a month ago from a lecture by John Conway on Faulhaber’s triangle. It gets kind of technical, but this pattern of multiplying by a fraction gotten by adding one to the numerator and denominator of the preceding fraction persists.

But in this case, to go from n(n+1)/2 to (n+1)(n+2)/2 you obviously have to multiply by (n+2)/n.

Oh, and as to why the sum of the first n integers is (n)*(n+1)/2:

Suppose I want to add the numbers 1+2+3+4+5+6+7+8+9+10

I can rewrite this as (1+10) + (2+9) + (3+8) + (4+7) + (5+6) you will note that all of the pairs add to 11, and that there are 5 of them so the total is 55.

In general for any even number n, I can do the same thing 1+2+…n= (1+n)+(2+(n-1))+(3+(n-2))+…(n/2+(n/2+1)). All of the pairs add up to (n+1) and there are n/2 of them so the total is (n+1)*n/2.

If n is an odd number, you can just start counting from 0. 0+1+…n=(0+n)+(1+(n-1))+(2+(n-2))+…+((n-1)/2+(n+1)/2). All add up to n and there are (n+1)/2 such pairs so you again get n*(n+1)/2 as the total.

I think geometrically is the nicest way to think of this.

Let’s call your n’th sum Triangle(n)

Now, there’s also a shape Rectangle(n) which you can make by duplicating Triangle(n), spinning the second one round and plonking it on top. Lets use 5 as the example:


(I made the second triangle out of circles not crosses for clarity)

At this point it’s really obvious why the formula for Triangle(n) is n(n+1)/2. It’s half of Rectangle(n) which is n(n+1)

So each line in the OP’s multiplication sum is something of the form:

Triangle(n) * {something} = Triangle(n+1)

We could do that with rectangles instead …

Rectangle(n) * {something} = Rectangle(n+1)

Gonna be the same ‘something’ in both cases, clearly, since we’ve simply multiplied both sides by two.

What is the ‘something’? Well, looking at, say,** Rectangle(5)** and Rectangle(6) it’s not immediately obvious…




How do you turn the first total into the second by multiplication? Looks tricky. After all, we’ve *added *one both horizontally and vertically - seems like figuring out a *multiplier *in the general case wouldn’t be that easy.

Oh, unless you rotate the first rectangle 90 degrees


  • {something}


Now all the lines have the same width and it’s obvious what we do - divide by the height of the first figure (5), multiply by the height of the second (7)

Rectangle(5) * (7/5) = Rectangle(6)

Or more generally…

Rectangle(n) * (n+2/n) = Rectangle(n+1)

and therefore of course

Triangle(n) * (n+2/n) = Triangle(n+1)

And we’re back to the OP…