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Video: Spotting Patterns in Visual Representations of Problems to Derive Formulae

Using the example of triangle numbers, this video gives a detailed explanation of the process of using diagrams and visualisations to help derive, validate, and use mathematical formulae.

08:52

Video Transcript

In this video, we’re gonna use an ancient method to analyse triangle numbers and derive a formula which enables us to work out the value of any triangle number. But using visual representations of problems can be a great way of understanding them and finding ways to produce algebraic formulae.

It’s a technique that the Ancient Greeks used to use using arrangements of pebbles or dots to represent problems. Imagine creating a display in a shop by stacking cans on top of each other in a specific plan. The simplest and, let’s face it, the most boring display would be one can on its own like this.

If we then add another row to the display following the rule that each row contains one more cans than the previous, we get these displays. For two, we’ve got two on the bottom row and then one on the top; for three, we’ve added a line of three cans to the bottom; and for four, we’ve added a line of four cans to the bottom.

So in the one row display, there’s obviously just one can in it. For the two row display, you’ve got one can on the top, two cans on the bottom, so that makes a total of three cans. In the three row display, we’ve added a row of three cans of the bottom of that so we’ve got six cans. And in the four row display, we’ve added a row of four to the bottom of that, which gives us a total of ten cans.

Now because the patterns made by these cans in the display make triangles, we call these numbers, one, three, six, ten and so on, triangle numbers. So the fifth triangle number would be the number of cans that you would get in a display following this rule, if you had five cans in the bottom row and the display had five rows. So that’s basically one plus two plus three plus four plus five, which makes fifteen.

And once you make the plan, you don’t actually even need to draw the cans anymore. Now what if you wanted to know how many cans would be in a display with fifty rows? We’d need to work out the value of the fiftieth triangle number.

So to work that out, we need to add up all the numbers from one to fifty. Now we could do that on a calculator or even in our heads. Or we could use Gauss’s famous method and we can spot that one plus fifty is fifty-one and two plus forty-nine is fifty-one and three plus forty-eight is fifty-one. And in that way we would then have up to twenty-five and twenty-six make fifty-one, so we’d have twenty-five lots of fifty-one, which is one thousand two hundred and seventy-five.

Now that’s great, but it took me a long time to write down all those numbers and it took a long time to do the calculation. Wouldn’t it be nice if we had a nice simple formula for working out the 𝑛th term or the 𝑛th triangle number? Well that’s what we’re gonna try and do now, come up with a formula for finding the 𝑛th triangle number.

Let’s go back to doing diagrams. And instead of piling cans on top of each other, we’re gonna shuffle everything to the left of the display so it makes a triangle which is a right-angled triangle that looks like this. So representing our triangle numbers this way, we can easily see that the first triangle number is one, the second triangle number is three, the third is six, and the fourth is ten, and so on.

So I can draw the pattern for the fifth triangle number, and I could just count them up. Then I can see that I’ve got fifteen dots, but what am I actually gonna do is I’m gonna replicate that pattern.

So I’ve now got two lots of the same triangle. If I was to count up all the dots in those two triangles, I would have twice as many as I needed; I’ll have thirty dots. But what I’m gonna do now is actually take the second pattern here and I’m gonna turn it round, so it looks like this, and then I’m gonna move it over to the other pattern and line it up like this. So now it makes a rectangle which makes it really really easy to count. So the rectangle is five dots wide. And because of the way that I took the original triangle out of a copy of the triangle and I’ve lined the copy triangle up like this, I’ve now got five dots high in the rectangle plus the other dot from the second copy of the triangle, so I’ve got six dots high.

So the number of dots that rectangle is five times five plus one, which is five times six, which is thirty.

Now remember, we said that’s twice as many dots as we needed. So if I halve that number, that’s gonna tell me the number of dots that I had in my original triangle. So just going back to our formula, our dot- rectangle had five dots wide and five plus one dots high. That gave us too many. So if we halve that, it’s a half times five times five plus one, which gives us fifteen dots.

Let’s try and generalise this to a formula for any triangle number. So let’s say we’ve got the 𝑛th triangle number, that means there would be 𝑛 dots across and 𝑛 dots up in a triangle representation of the problem. Now I know that there are six dots across and six dots up. And let’s imagine there could be seven, there could be a hundred, there could be a billion dots across and a billion dots up. That’s the basic pattern that it would follow. Now if I duplicate that triangle and then turn it round and slot in above, I’m gonna get a rectangle which has got 𝑛 dots wide and 𝑛 plus one dots up. And that contains twice as many dots as I need for my 𝑛th triangle number.

So to work out how many dots are in that rectangle, it’s width times height. That 𝑛 is the width, 𝑛 dots wide, times 𝑛 plus one; that’s the number of dots high it is. But remember, that’s twice as many dots as we need for our 𝑛th triangle number, so the 𝑛th triangle number is half of that amount.

And so we have a general formula for quickly working out the 𝑛th triangle number without having to add up one plus two plus three plus four plus five plus six plus seven and so on. So to work out the fiftieth triangle number, we would say that 𝑛 is equal to fifty. So plugging in that value 𝑛 equals fifty into our formula, we’ve got a half times 𝑛 is fifty times 𝑛 plus one is fifty plus one.

Well a half times fifty is twenty-five and fifty plus one is fifty-one, so we’ve got twenty-five times fifty-one, which is one thousand two hundred and seventy-five, which luckily is exactly the same number that we had before.

So now we’ve got our formula for triangle number. We can work out any triangle number. Let’s have a go at these three questions. Work out the tenth triangle number. Well in this case 𝑛 is ten, so we have to use a value of ten in our formula for 𝑛. So our tenth triangle number is gonna be the a half times ten times ten plus one, which is five times eleven, which is fifty-five. So the tenth triangle number, is fifty-five. You have a go at b) and c) now, so pause the video and then come back and I’ll show you the answers.

So for part b), the hundredth triangle number 𝑛 will be equal to a hundred. And plugging that into our formula, the hundredth triangle number is a half times a hundred times a hundred plus one. So that’s fifty times a hundred and one, which is five thousand and fifty.

And the two hundred and seventy-fifth triangle number, 𝑛 will be equal to two hundred and seventy-five. So that’s a half times two hundred and seventy-five times two hundred and seventy-five plus one.

Now because we started off with an odd number for 𝑛 this time, it’s gonna be easier to do a half times two hundred and seventy-six. And that will give us a hundred and thirty-eight times two hundred and seventy-five, which is thirty-seven thousand nine hundred and fifty. And I’m sure you’ll think that that’s a lot easier, even though the numbers aren’t entirely easy, certainly a lot easier than adding up one plus two plus three plus four all the way up to two hundred and seventy-five. That will take absolutely ages.

So sometimes drawing diagrams and playing around with those representations and trying to spot the patterns that you can see in those visual representations of the problem can help us to come up with a formula much more quickly. In this case, we’ve worked out the formula for the 𝑛th triangle number and that is a half times 𝑛 times 𝑛 plus one.

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