GCSE Maths: Quadratic Sequences Explained

and welcome to this video on the quadratic nth term. I'm going to go through a bunch of the exam questions that have come up in the last four years. Go through every single one and explain it to you in a method that I prefer that I think it relies less on memory and it kind of makes more mathematical sense which will make it easier for you to actually remember in the exam and get those full marks. But before we get started, I just want to let you guys know if you look at the pinned comment or the link in the description, I have some spaces in my group classes for GCSE and Alevel maths tuition. So, if you're interested, you can book your free one-on-one trial again with the link in the pin comment or in the description. But let's get into the first question. So, a quadratic sequence, we need to find the nth term. It's a super super common question, comes up most years. The first thing all you need to do, and I'm going to write out the sequence just one more time to make it a bit clearer, is you started the same way as you do with the linear sequence by just finding out what the common difference is. Now, what you're going to notice is the common difference is not the same each time. In fact, it changes. So, it's going up by well, you can see it's going up by four each time. Okay. Now, when we think of the term quadratic, I'm hoping that you're thinking of an equation that looks like this. So, as an example, you might have 2x^2 + 3x + 4 or something like that. That's what a quadratic equation looks like. That means we expect the quadratic sequence to have an equation that looks like this. So it should have an x squ and it should well actually just an x squ. It doesn't have to have the x. It doesn't have to have the c. So when we're doing this, we're going to write out that our equation is going to look something like this. Now the only thing it has to have is n squ. It might have a bn. It might have a c as well. Now the next thing you're going to do is you're going to find the second difference of this. So you're going to work out what the difference is going up by in each time. Now here's the only bit you need to remember. The only bit you need to remember from this whole method, which is that a is always half of that common difference all the time. So if it was -6, half it is -3, whatever. We're going to go through some more examples anyway. So a is always going to equal two in this case. It's going to be half of that common difference. So that means we now know our equation looks like this. 2 n ^ 2 plus bn + c. But how do we find bn and c? Well to do that we need to think about what n actually means. So what I'm going to do is I'm going to drag this down to give us a little bit more space. So I'm just going to move this about a bit. Drag this down. and then we can have a little bit more space to talk about it and think about what N represents. N represents what position you're in in the sequence. So for example, N is one gives us the first term of the sequence. It should give us five. n is two should give us the second term of the sequence which is 11. n is three should give us the third term of the sequence which is 21 and so on. So how can we use that to help us? Well, it's actually exactly what I just said. When n equals 1, we should get an answer of five. So when n equals 1, we should get five. So let's actually sub that in. We get 2 1 2 + b * 1 + c = 5. So if we just expand that quickly, spoilers, we get 2 + b + c = 5, which gives us b + c = 3. And that's our first equation. We're going to do the same thing for n equals 2. I'm going to explain why I picked those two numbers in a second. It will make it a bit more sense. So again, we're going to sub it in. We get 2 * 2^ 2 + b * 2 + c. And what should that equal? It should equal 11, the second term in the sequence. So 2 ^ 2 is 4 * 2 gives me 8 + 2 b + c = 11. So we get 2 b + c = 3. Now I'm going to show you why we picked those two values because we can actually solve that. Now it's a simultaneous equation. Now I know you're thinking simultaneous equation that sounds difficult. But take a look at how quick we can solve these two equations. I'm just going to write them on top of each other for you. So I've written them on top of each other. Look how easy this is to solve. The number in front of C is the same already in the equation. So, if I subtract these two equations, I get B equals 0. I've already worked out what B is. And now I know that B is zero. I can just sub that straight back in, and I can get 0 + C = 3. So, C is equal to 3. We solved that in about 10 seconds. That's the power of this method. It's very, very quick, but at the same time, it is it's based on the maths you already know. you're not really learning anything new. So it means our answer is 2 n^ 2 b is 0. So that's 0 + 3. So 2 n ^ 2 + 3 for that answer. So why did I pick n is 1 and n equals 2. Well, if you have a look for a second, the reason why this was so quick to solve is because this 2b and this b are the exact, you know, they're only one away. When we subtract them, we're just left with b. That's because I picked n is 2 and n is one. If you always pick n is one and n is two, you can solve it this quickly every single time. And I mean every single time bar none. You always solve it that quickly, which is really, really nice. Now, of course, you should check your answer by subbing in different values. So, as an example, if I sub in n is 1, I should get five. So, 1 2 is 1 * 2 is 2 + 3 is 5. If I sub n is 2, I should get 11. 2 is 4 * 2 is 8 + 3 is 11. So, I have got the right answer. I'm going to give you another one. I'm going to do this one and I want you guys to do the one after this. Okay? So, I'm going to let you pause the video in a second. If you really feel confident, you can pause it now, but I'm going to talk through this one immediately. So, when I write out the sequence, what was the first thing I did? I found the difference between these. So, that goes up by 11. That goes up by 17. That goes up by 23. that goes up by 29. And then we find the common difference between that. So that's six and it should be the same. If it's not the same, it means it's not a quadratic sequence. So perhaps you've done something wrong slightly. Why was that important? Can you remember? If you remember, we have a n^ squ plus bn plus c. The a is always half of this common difference. So straight away we know a is equal to 3. Super super useful. So our equation so far is 3 n^ 2 + bn + c. Now what did we do next? So that's the first major step. What was the second major step? The simultaneous equation. So what we did if you remember we picked n is 1 and n is two. Because if we pick one and two it makes the simultaneous equation really really easy to solve. So when n equals 1 we should get 10 because the first term is 10. So let's do that. n= 1, we should get 3 1 2 + b 1 + c = 10. So that's 3 + b + c = 10. So b + c = 7. That's my first equation. Do the same thing with n= 2. If n equals 2, we should get 21 because that's the second term in the sequence. So if we sub that into again back into our equation we've gone so far 2 + b 2 + c then 2 ^ 2 is 4 * 3 is 12 2 b + c in fact I should write that equ= 21 = 21 so 2 b + c = 21 - 12 is 9. Now it's just solving it. How do we solve it? We write them on top of each other and it's a super super super super easy and simple and quick equation to solve. We just always subtract them and we always get B equals something. We get B equals 2 in this case. Then you sub it back into that bottom equation, you get 2 + C = 7, you get C= 5. How quick is that? It's always that quick as well. It doesn't matter what question you get, you will always get something that fast to solve, which is really, really nice. So our final equation, remember we know that B is equal to 2. Now we have C is equal to 5 is this. Our final answer is 3 N 2 + 2 N + 5. Done. If you'd like, you can check by just subbing in n= 1 and n= 2 just to see if you get that same sequence. Now, here's one I want you guys to do. So please do pause the video. Feel free to look back earlier in the video so you get a full example if you've forgotten how to do part of it. But again, through doing this, you'll actually kind of improve faster. Okay, that should be enough time. Let's go through it. So again, we write out that expression. What do we do next? You find the difference. That's going up by one. That's going up by three. That's going up by five. By the way, it can also be a negative number. It does not have to be a positive number. The difference can be negative. That goes up by plus two. that goes up by + two. That goes up by plus two. Why do we do that? Because a the number in front of n^ squ is always half of that difference which is one. It could also be a fraction. Okay, it can be. There's no reason why not. So means our equation so far is n^2 + bn + c. What do we do next? We sub in n= 1 and n= 2 so that we can solve it. So when n= 1, what do we get? we get 1^ 2 + b * 1 + c = -1. So we get 1 + b + c =1. So that's b + c = -2. Okay? Because you subtract one from both sides. Then we do the same thing with n= 2. When n= 2, we should get 2 ^ 2 + b * 2 + c = 0. Again, let's expand it. So, we have 2 ^ 2, which is 4 + 2 * b is 2 b + c = 0. So, we get 2 b + c = -4 because we subtract four from both sides. Then all we do is write them on top of each other, subtract them. Done. Super nice. So, oh man, spoilers. I need to remove this question. So, let's put them on top of each other and solve it. Again, hopefully you're just checking your work as you went along. Now, we subtract it. So, 2 B minus B is B. C - C is nothing. -4 - -2, be careful with that, is -2. Subbing that back into here, we have -2 + C = -2. Add two to both sides, C is zero. So, those are our two values. So, let's write this out. we have n ^ 2 - 2 n + 0. So hopefully you got that one right, but if not, I'm going to give you another chance. This is the last exam question that's come up. So again, we've done the last four or five years of exam questions. Have a go at this one. See if you can get it right. Okay, let's start going through this exam question again. What does it go up by each time? That's the first step in every single question. So that goes up by six. That goes up by 10. then 14. Uh that should be 18 and 22. Then we find the next difference. So as you can see it's going up by four each time. Which means what's a? a should be 4 / 2 which is 2. So that gives us so far 2 n^ 2 + bn + c. And now we just need to solve the quadrat uh the simultaneous equation by subbing at n= 1 and n = 2. So when n= 1 we get 2 1^ 2 + b 1 + c should equal -1 just from the question. 1 2 is 1 * 2 is 2. So 2 + b + c =1 minus 2 from both sides we get b + c is -3. That's our first equation. Then when n= 2 we get we should get five but before that we get 2 2 + b * 2 + c = 5 2 ^ 2 is 4 * 2 should be 8 + 2 b + c = 5. Subtract 8 from both sides. We get 2 b + c = -3. And now all we have to do is solve it by putting them on top of each other and subtracting as usual. So that gives us 2 b + c = -3 and on the bottom it's b + c = -3. Cool. Subtracting those two equations gives us b = -3 - -3 is 0, of course. And subbing that back in we get 0 + c = -3. So that means c is equal to -3. So our final answer is 2 n ^ 2. There's no b. Okay. And then minus 3. So again, you can get b or c being zero. You will never get a being zero because there's nothing you can divide by two to get zero except zero. So yeah, hopefully that makes a lot more sense. Once again, if you're interested in booking that one-on-one um kind of initial trial session with me, then do see the link in the pin comment or the description. But I'll see you in the next