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Video: Binomial Expansion

Lucy Murray

Learn how to use the numbers in a specific row of Pascal’s triangle, and the binomial theorem, in order to quickly expand parentheses in the form (𝑥 + 𝑦)^𝑛, also how to use 𝑛 choose 𝑟 term to help calculate the coefficients in a binomial expansion.

14:12

Video Transcript

Binomial Expansion There was a famous French mathematician called Blaise Pascal, and he came up with this pattern called Pascal’s triangle, named after himself. So if we start with one, this is our first row, so we’ll call this row 𝑛 equals zero.

Now underneath one, I’m gonna write two more ones. And we’ve got these two numbers by adding the two numbers above them, so looking at the first one, we’ve added one and zero together to get one. And then looking at the second one, the same thing. We’ve also added one and zero.

Now this line is our 𝑛 equals one line. Looking at the next line, we can see exactly which number’s gonna go underneath those two ones, because we’re gonna add them together and we’ll get two.

And then again on either side of the two, we’re adding zero on one, so we’re going to get one on either side of the two. This is the term 𝑛 equals two.

And moving on to the next line, adding two and one together twice gives us three. You should by now have worked out what’s gonna go on either side of the threes: one, good. So each time, we’re just adding the two numbers on top to give us the next number. So again moving down from our 𝑛 equals three to 𝑛 equals four.

So first of all, we’re gonna have a one on either side, and then one add three is four, three add three is six, and three add one is four.

This probably seems a bit strange right now, saying why are you just making a triangle and adding things above to go down rows? But you should be able to see that we could carry on going for infinity. That’s just up to row 𝑛; we could carry on going forever. Let me try and show you how we’re going to use it.

One of the ways we use it is by expanding brackets. So if I have 𝑥 plus 𝑦 all to the power of zero, we know that anything to the power of zero is just gonna be one. How about 𝑥 plus 𝑦 to the power of one? Well that’s gonna be 𝑥 plus 𝑦. I wouldn’t usually but on this occasion I’m going to write the coefficients of one.

Okay let’s try 𝑥 plus 𝑦 all squared. We know that’s the same as all of 𝑥 plus 𝑦 multiplied by all of 𝑥 plus 𝑦, so let’s do it on the side. Multiplying out using FOIL, we’ll do the first terms, so 𝑥 multiplied by 𝑥 gives us 𝑥 squared.

Outside is 𝑥 multiplied by 𝑦, giving us 𝑥𝑦. Inside which is 𝑦 multiplied by 𝑥, giving us 𝑥𝑦. And last which is 𝑦 multiplied by 𝑦, giving us 𝑦 squared.

So we can collect those like terms giving us 𝑥 squared plus two 𝑥𝑦 plus 𝑦 squared. And then if we write that down back where we originally were. We’ll do one more multiplication; we’re gonna have 𝑥 plus 𝑦 all cubed. So this we know is all of 𝑥 plus 𝑦 multiplied by all of 𝑥 plus 𝑦 multiplied by all of 𝑥 plus 𝑦, but we’ve already done 𝑥 plus 𝑦 all squared, so we’re just gonna take that and multiply it by 𝑥 plus 𝑦 again.

So to do this, we’re going to take every term in the first set of parentheses and multiply them by 𝑥. And then we’re going to add on to that everything in the first set of parentheses multiplied by 𝑦.

Okay now and if we collect the like terms, we can see we’ve got two 𝑥 squared 𝑦, and then we’ve got another 𝑥 squared 𝑦, so we’ll have three 𝑥 squared 𝑦s. We’ve got one 𝑥𝑦 squared and another two 𝑥𝑦 squared. So adding that together, we’ll get three 𝑥𝑦 squared, and then finally a 𝑦 cubed.

Okay so then let’s write this back where the others are. So hopefully, you should have started to notice something. If we look at all the numbers in blue for our expansions and the numbers in the Pascal’s triangle, you should be able to see that they’re exactly the same. And if you look at the powers for each of the expansions, the first one is to the power of zero. Remember I said that’s where 𝑛 is equal to zero, that top line? And then the next one we’ve got the power of expansion is one. That’s the line of 𝑛 equals one, and so on and so forth, so that 𝑛 actually talks about what we’re putting our parentheses to the power of. Now we’re gonna have a go. I’m gonna show you what the binomial expansion theorem is, and we can have a look at how we would apply that and apply our knowledge now of Pascal’s triangle to help us expand really large brackets without having to go through all that hassle we just did then.

So the binomial theorem is, 𝑎 plus 𝑏 all to the power of 𝑛 is equal to the sum of from 𝑘 equals zero to 𝑛 of 𝑛 choose 𝑘 multiplied by 𝑎 to the power of 𝑛 minus 𝑘 multiplied by 𝑏 to the power of 𝑘.

This seems little bit confusing. So let’s actually show what that means. So it means 𝑛 choose zero multiplied by 𝑎 to the power of 𝑛 plus 𝑛 choose one multiplied by 𝑎 to the power of 𝑛 minus one multiplied by 𝑏 to the power of one plus every single term in between and then 𝑛 choose 𝑛 minus one all multiplied by 𝑎 to the power of one all multiplied by 𝑏 to the power on 𝑛 minus one plus 𝑛 choose 𝑛 multiplied by 𝑏 to the power of 𝑛.

So in our calculators, we can put in 𝑛 choose any number, and it’s usually a button that looks like this, depending on which calculator you have. But if we don’t have a calculator, then we can either use Pascal’s triangle to give us the 𝑛 choose whatever we’re looking for or we can use factorials, where 𝑛 choose 𝑘 is equal to 𝑛 factorial all divided by all of 𝑛 minus 𝑘 all factorial multiplied by 𝑘 factorial. In this video, we’re going to focus on what would happen if we use our calculator and also Pascal’s triangle, but you can apply this, what we’ve just shown there, where 𝑛 choose 𝑘 with factorials, to any one of the examples we do.

Expand 𝑥 plus two 𝑦 all to the power of five using binomial expansion. Okay so let’s apply the expansion; we’ll have five choose zero multiplied by 𝑥 to the power of five plus five choose one multiplied by 𝑥 to the power of four multiplied by two 𝑦 to the power of one.

So we can see here our powers in 𝑥 are going to decrease by one each time, and our powers in 𝑦 are going to increase by one each time. So then the next time, we’ll have five choose two multiplied by 𝑥 cubed multiplied by two 𝑦 all squared plus five choose three multiplied by 𝑥 squared multiplied by two 𝑦 all cubed.

We’ll have to go on another line for this. I always end up doing them with binomial expansion. That’s five choose four 𝑥 to the power of one so 𝑥, all multiplied by two 𝑦 to the power of four.

And then our final term five choose five, now 𝑥 because it’ll be 𝑥 to power of zero, and then multiplied by two 𝑦 all to the power of five.

So before we start calculating anything, let’s just have a look at what we’ve done. We can see that each time we’ve got 𝑛 is equal to five; then we’re choosing a number that’s increasing each time until we get to five. We can see the 𝑥 powers are decreasing by one each time and the 𝑦 powers are increasing by one each time. Now having a go at calculating it, you can put in your calculator like I said using the button five choose zero, five choose one, et cetera. I’m just gonna grab the fifth line from Pascal’s triangle, which if you remember the fourth line was one four six four one, but then adding the two numbers above, we will get one, five, ten, ten, five, and one.

These will be the values of all of our 𝑛 choose 𝑟s individually. So first, we have one multiplied by 𝑥 to the power of five, so that’s just 𝑥 to the power of five. Then five choose one which is five. Multiplying that by two 𝑥 to the power of four and 𝑦.

Then we can see the five choose two will be ten. We’re gonna multiply that by two squared, two squared we know is four, and 𝑥 cubed and 𝑦 squared, and five choose three we’ve got another ten.

Multiplying that by two cubed, two cubed is eight, and then by 𝑥 squared and 𝑦 cubed. And five choose four we can see is five. Multiplying that by two to the power of four which is sixteen, and 𝑥 and 𝑦 to the power of four.

Then for our last term, we can see five choose five is one, and we’re multiplying that by two to the power of five, which is thirty-two 𝑦 to the power of five.

Okay and then if we find out all the numbers here, we’ve got 𝑥 to the power of five plus ten 𝑥 to the power of four 𝑦 plus forty 𝑥 cubed 𝑦 squared plus eighty 𝑥 squared 𝑦 cubed plus eighty 𝑥𝑦 to the power of four plus thirty-two 𝑦 to the power of five. And we have expanded that bracket using binomial expansion, and that is a lot quicker than if we’d have had to do each of those multiplications out of the brackets individually. But what if in a question we’re not asked for the whole thing? What if we’re just asked for one term?

What is the fourth term in the expansion two 𝑥 minus three all to the power of eight? So first of all, think about the things we need to do. First of all, we need to do 𝑛 choose sum 𝑟. We know the 𝑛 will be eight, and the 𝑟 will be three.

As if we remember that 𝑟 starts off on zero, so the fourth term would go zero one two three, so three would be the fourth term. This then is able to tell us the power of two 𝑥; we’re gonna put two 𝑥 to the power of five because it’s two 𝑥 to the power of eight minus three.

And then we know it’s going to be negative three all to the power of three, as that is our 𝑟. So if you put eight choose three into a calculator, you get an answer of fifty-six.

You’re gonna multiply that by two to the power of five, which we know is thirty-two, and negative three to the power of three, which we know is negative twenty-seven. Of course, don’t forget 𝑥 to the power of five.

So then if we multiply these altogether in our calculators, we get a rather large number of negative forty-eight thousand three hundred and eighty-four 𝑥 to the power of five.

So there we have it. To find individual terms, we need to be sure of the 𝑛 and the 𝑟. To find the 𝑛, it’s just gonna be whatever power the bracket is to. So in this case, it is eight. And finally 𝑟 is gonna be one less than the term number we want.

So in summary, we have used Pascal’s triangle and binomial theorem to help us carry out binomial expansion of some very large brackets. And binomial theorem, this formula, is specifically helpful for if we want to find just individual terms because if we ignore the sum that’s exactly what we’ll use.