Triangular Day: Tripled

I don't know if you noticed, but today is a triangular day (and a week away from our hundredth post). It is the sixth, and six is a triangular number. We haven't really tried to prove stuff in a little while, so I wanted to take the time and do that today.

First off, let me tell you what we are going to do. We will prove that any triangular number tripled plus the next consecutive triangular number equals a triangular number.

That was a lot of information there. Let's just do the pattern so you can see better.

1) 3(1) + 3 = 6
2) 3(3) + 6 = 15
3) 3(6) + 10 = 28
4) 3(10) + 15 = 45
5) 3(15) + 21 = 66
6) 3(21) + 28 = 91
7) 3(28) + 36 = 120

That is pretty cool. But proving it is even better. And just like lots before this one, we can prove it with algebra.

Did you notice any pattern in the numbers coming out? Look at the positions of the numbers in the sequence rather than the number itself.

1 ––> 3
2 ––> 5
3 ––> 7
4 ––> 9

It consistently works out to the following:

3T(n) + T(n+1) = T(2n + 1)

(Remember that T(n) means the nth triangular number).

Using the triangular number formula n(n+1)/2, how do we write these three terms?

3T(n) = 3n(n+1)/2
T(n+1) = (n+1)(n+2)/2
T(2n+1) = (2n+1)(2n+2)/2

So, we can write it as:

3n(n+1)/2 + (n+1)(n+2)/2 = (2n+1)(2n+2)/2

Since I don't love working with fractions, let's multiply both sides by two and get rid of any denominators.

3n(n+1) + (n+1)(n+2) = (2n+1)(2n+2)

Now, let's simplify all of these factored terms.

3n^2 + 3n + n^2 + 2 = 4n^2 + 3n + 2
4n^2 + 3n + 2 = 4n^2 + 3n + 2

Both sides simplify to the same thing, thereby proving the pattern to be correct.

We can also prove this geometrically. There were tiles in first grade that we would play with that had different shapes, like squares, triangles, trapezoids, hexagons, etc.

We always would try to create a really big triangle. A shortcut we used was instead of making the second row with three triangle pieces (two facing up and one facing down), we would just use a trapezoid. If we were to take a three-dotted triangle and turn it into a trapezoid, we would just do this:

•  •  •  •  •
  •  •  •  •

You can see the triangles better like this:


•  •    •    •  •
  •    •  •    •


This relates perfectly to this pattern. We are adding to a triangle (the second number) to get a bigger triangle (the total). And what exactly are we adding? Three triangles, which makes a trapezoid.

Let's use some logic here. The triangle we are adding onto has a length of n+1, as it is T(n+1). So, the trapezoid should have a smaller base of n+2.

The full triangle at the end should have a side length of 2n+1, meaning that is going to be the bigger base of the trapezoid.

To fully prove it, we think about the actual construction of the trapezoid. Remember how I said that two triangles would point up and one would point down?

For the smaller side, the two triangles that point towards it each add one dot to it. The one that points the other way adds n dots to it. You can see that better if you look at the diagram above.

1 + 1 + n = n + 2

For the bigger side, we have only one triangle pointing towards it with a side of one. The others add n dots to it.

1 + n + n = 2n + 1

And there is the proof. I found this fascinating because of the pattern itself, as well as the fact that you can prove it many ways, which is common with triangular numbers.