Loading...
Thank you for joining me for today's lesson.
My name is Ms. Davies and I'm gonna be guiding you as you explore some of these new and exciting sequences that we're looking at today.
Make sure that you've got everything you need before you start watching this video.
It's always a good idea to have a pen and paper so that you can jot things down and explore things in your own time.
Let's get started then.
Welcome to this lesson on sequence notation.
Today, we're gonna look at some new notation that we can apply to sequences.
If you would like a reminder of what the nth term is, what a term-to-term rule is or what a Fibonacci sequence is, pause video and look through those now.
So we're gonna start then by looking at this new sequence notation.
So a linear sequence starts 3, 7, 11, 15, 19.
How could you describe this sequence? Pause the video.
What can you come up with? Right, there's lots of ways to describe the sequence.
You may have said, successive terms have a common difference of 4.
You may have said it starts on 3 and has a term-to-term rule +4.
You may have said each term is one less than the 4 times table, or you might have used the nth term rule of 4n - 1.
Why is that second description, it starts on 3 and has a term-to-term rule +4 better than the one above it? Well, the main reason is because it includes the first term.
Now there's lots of different sequences that have a common difference of four.
If I want to refer to this one, I need to say that it starts on three and has a common difference of four.
And that's where nth term rules can come in handy.
nth term rules are a really good way of describing some sequences.
They uniquely describe a sequence without needing extra information.
They can be used to easily find any term given its term number and they can be used to form equations to solve certain problems. These are all things you've probably done before.
Can you think of any disadvantages of using an nth term rule to describe a sequence? I wonder if you thought of any of these.
They can be quite tricky to work out for some types of sequences.
They do not show how to easily generate the next term.
So if you have four terms, you want the next term, they're not always easy to use to just find the next term.
For some sequences, there's not an obvious mathematical relationship between the term number and the term.
So in contrast, term-to-term rules are a good way to describe the relationship between successive terms in a sequence.
This means they can easily find the next term given the previous one.
So we're gonna use new sequence notation to efficiently write term-to-term rules.
Okay, so our first term, we're gonna write as u subscript 1.
The lowercase u is used to show that we are talking about a term in a sequence.
The subscript 1 tells us we are referring to the first term.
You might hear it said as u1.
It's important to know that that 1 is in subscript.
The second term would be u subscript 2 and the third, u subscript 3.
How do you think we would refer to the 10th term? Maybe have a go at writing it down.
We'd write it as u with a subscript 10.
So let's return to our previous sequence, u1 would be 3, u2 would be 7, u3 would be 11, u4 would be 15, and u5 would be 19.
Now we're still need to define a rule, otherwise, there's no guarantee that the sequence will continue with a linear pattern.
We want to define a linear rule.
So for this sequence, any term is the previous term add 4.
Well this means we need a way to denote any term.
What we're gonna use is a u subscript n where n is the term number.
So if I use U subscript n, I'm referring to any possible term number.
If un is any given term in the sequence, we can call the next term, u with a subscript n + 1.
It's important the whole of n + 1 is subscript.
This is why notation is better than saying things in words because it's really clear what we mean by the notation.
It's quite hard to say it in words.
Where possible, make sure you are looking at the way things are written, not just how they're being spoken so you don't get confused.
So let's say, the fourth term, n would be 4.
So un would equal u4.
The next term, un + 1 would be equal to u4 + 1 or u5.
Well, that makes sense, doesn't it? As we're adding one to the number in the subscript, we're getting the next term.
So we can use this notation to write our term-to-term rule.
So u subscript to n + 1 equals un + 4.
The next term in the sequence is equal to the current term add 4.
We also need to write the value of the first term, otherwise, this sequence could start on any value.
We said this at the beginning with term-to-term rules.
So we need to write where u1 equals 3.
A different sequence starts on 10 and subtracts two each time.
How do you think we could write this with our new sequence notation? You might wanna try this.
So we're gonna have u with a subscript of n + 1 equals u subscript n - 2.
Where u1 equals 10.
Don't forget you need that first term.
So what this says is the next term in the sequence is equal to the current term subtract two or the first term is 10.
Quick check then for the linear sequence 6, 11, 16, 21, what is the correct notation for the term-to-term rule? Well done if you picked B.
A is no good to us 'cause it doesn't give us the first term.
There's all sorts of sequences where you add five to get the next term, we need to know where we're starting.
And we can see from our sequence that we're adding five each time and we're starting on six.
How about this one.
the linear sequence, 17, 19, 21, 23.
What's the correct notation? Well done if you picked C.
Our term-to-term rule, we're adding two each time.
It's really important that we're writing u of n add 2.
So we should have u with a subscript of n + 1 equals u with a subscript of n + 2 where u of 1 equals 17.
Just make sure you are happy with that form.
So this notation can help us denote sequences with trickier term-to-term rules.
So this sequence starts on five and the next term is double the current term.
We can write this as u with a subscript of n + 1 equals 2 lots of u with a subscript n.
un plus 1 equals 2un where u1 equals 5.
As long as there is a way of generating the next term from the previous term using a mathematical rule, we can write this using our new notation.
Jun says, "I've come up with this interesting sequence where you double the previous term and subtract 1." We've got 2, 3, 5, 9.
Luca says, "Your sequence is not arithmetic, geometric or quadratic.
I have no idea how to write an nth term rule." And finding an nth term rule would be tricky.
So how could we write this using our new notation? Right, well using a term-to-term rule is a lot easier than an nth term rule for this example because we know that each term is double the previous term then subtract one.
So we can write u subscript n + 1 equals 2u subscript n - 1 where u1 equals 2 or the next term is two lots of the current term subtract one, where the first term is two.
We can generate the first few terms of a sequence given this notation.
So let's look at the first three terms of this sequence.
So we've got un + 1 equals 3 lots of un plus 5 where u1 is 1, so the first term is one.
u of 2 is gonna be 3 lots of u1 plus 5 or 3 lots of 1 plus 5, which is 8, u of 3, there's 3 lots of u2 plus 5 or 3 lots of 8 plus 5, which gives us 29.
So we've got 1, 8, 29.
Each term is three times the previous term then add five.
Now what are the disadvantages of writing a sequence this way? Is that it does not give an easy way to generate any term.
An nth term does, but this way of writing a sequence doesn't.
I want to find the 10th term in this sequence.
So I can say that u10 equals u9 plus 5, but I can only do that if I know the 9th term.
Well, the 9th term is u8 plus 5, but I need the 8th term to find the 9th term.
So we'd need to start on the first term and generate each term until we got u10.
So u1 equals 4, u2 equals 4 plus 5 which is 9, u3 equals 14, u4 equals 19, and we could keep going until we've got the 10th term.
And that's where an nth term rule would have the advantage over using this notation.
Right, quick check.
I'd like you to match the term-to-term rule with the correct notation.
Off you go.
A is gonna match up with H.
The next term is the current term add three, starting with one.
B is gonna match up with E.
The next term is the current term add one when the first term is three.
And notice the difference between u with a subscript of n + 1 and u with a subscript of n or plus one.
They're completely different things so make sure you're confident with that.
C is gonna match up with F.
The next term is three lots of the current term starting on one.
And finally, D is gonna match up with G.
We've got 2 lots of un plus 1, which means double the current term and add one.
All right, this time working the other way around, can you match the sequences notation to the correct first four terms? So the first one, we're saying the next term is the current term add three starting on one, so that's F.
B, you've got the next term it's the current term add one starting on three, that's E.
C, we've got the next term is three lots of the current term starting on one, and D, we've got the next term is two lots of the current term add one starting on three, so that's G.
So this notation might look a little confusing at first 'cause you're not used to using the subscript.
However, this is just us formally describing term-to-term rules, which you've been looking at before.
So hopefully, you're starting to see that this notation is actually quite straightforward.
Time to have a practise.
I'd like you to write each term-to-term rule using our new notation.
Don't forget, you also need to give us the first term.
Give that a go.
Well done.
For this second set, I've given you the notation.
Can you generate the first four terms? Off you go.
And a little bit more of a challenge this time.
I've given you some information about sequence.
Can you write it using our new notation? For D, think about what the terms in the sequence are gonna be before you turn it into a term-to-term rule.
Let's have a look then.
So pause the video and check you've got these correct.
Pay particular attention to when you've got subscript and when you haven't.
Don't forget to include your first term as well.
For E, there's several ways that you could have written that, so just have a look at those options.
For question two, we've got first term of seven and a term-to-term rule of add eight.
So you have 7, 15, 23, 31.
For B, first term 22, a term-to-term rule of subtract 10, 22, 12, 2, -8.
For C, we've got a first term of four and a term-to-term rule of multiply by five, we've got 4, 20, 100, 500.
And for D, we've got first term three and then we're multiplying by 10 then subtracting 10.
So we get 3, 20, 190 1,890.
For E, we've put some brackets in this time, so we're starting on three, but this time we're subtracting 10 before multiplying by 10.
We get 3, -70, -800 and -8,100.
For F, we're starting on -3 and we're multiplying each term by -1 to get the next term.
So -3, 3, -3, 3.
And G, we're starting on 32 and we're dividing each term by four to get the next term.
32, 8, 2, 1/2.
Then our challenge.
So for A, you should have u subscript n + 1 equals u subscript n + 6 where u1 is 5.
For B, you should have u subscript n + 1 equals 4 un where u1 equals 3.
For C, first, we need to work out our multiplier.
So 12 divided by 18 is 2/3 so is 8 divided by 12.
So u subscript n + 1 equals 2/3 un where u1 equals 18.
For D, it helped to write out the first four terms of the sequence, spot the term-to-term rule and then put it in notation.
Just check you've got those correct.
And there you have it.
You've learned this new notation.
We're now gonna have a play around with what we can use this for.
Aisha says, "My sequence can be written as u subscript n + 1 equals 2u subscript n + 3.
Why can we not generate Aisha's sequence? Right, we don't know the first term.
We could pick any number, double it and add three to create a sequence but would not know if it was Aisha's sequence.
She says, "My sequence contains the number 17." Can we now generate her sequence? No, of course not.
We need to know which term in the sequence 17 is.
If 17 was the first term, then we'd have 17 double and three gives us 37, double and add three gives us 77.
Aisha says, "Actually it's u3 which is 17." For this type of sequence, we only need to know one term and then we can calculate the others.
So normally, it's helpful if you're given the first term.
If you're given a different term, we can still work it out.
So how could we calculate u2? Well u3 is 2 lots of u2 plus 3.
So that means u3 minus 3 equals 2 lots of u2, so u3 minus 3 all over 2 equals u2.
So what we have there is we have a rule for finding u2, if we know u3.
Now we do know u3 so let's put our value in.
So we've got 17 minus 3 over 2.
So u2 equals 7.
Right, what is u1? Or u1 is gonna be u2 minus 3 over 2 which is 7 minus 3 over 2, which is 2.
So we have the sequence 2, 7, 17, then we could have 37 and 77.
You might wanna check that that follows the rule double and add three.
Right, if we have a relationship showing us how un and un + 1 are connected, then we can use one to find the other.
For example, look at this relationship.
Now we could rearrange this to work out how you could find un if you knew un + 1.
All we need to do is rearrange like an equation.
So we divide both sides by two.
We've got u subscript n + 1 divided by 2 equals u subscript n + 1.
Then we can subtract one from both sides.
So u subscript n + 1 divided by 2 minus 1 equals u subscript n.
We can now use this to find a previous term if we know the current term.
So u4 divided by 2 minus 1 would give us u3.
So u3 is 10.
Have a go then at calculating u1 and u2.
Well, u2 is gonna be easiest to calculate first.
So u2 is gonna be u3 divided by 2 minus 1 or 10 divided by 2 minus 1, and then u1 is gonna be u2 divided by 2 minus 1 or u1 will be 1.
So we've got 1, 4, 10, 22.
Quick check.
Have a read of this sequence notation.
Which of these are the first five terms? Well done if you picked A.
Notice that it's the third term, which is six.
Our term-to-term rule is add two.
So to get the next term, we should add two and then add two again.
So that's gonna give us 2, 4, 6, 8, 10.
A sequence can be denoted this way.
What is the value of the first term? Give this a go.
Right, we should have two.
If we think about rearranging this, if we add five to both sides and then divide by five, we know that un is gonna be the same as u subscript n + 1 plus 5 all over 5.
And then you can just put in u2.
If we do 5 plus 5 over 5, that'll tell us u1.
Now Fibonacci sequences can be expressed using this notation.
This is quite interesting.
Let's use the Fibonacci sequence 1, 1, 2, 3, 5, 8.
Describe in words how any term in the sequence is generated.
Give this a go.
Right, each term is generated by adding the two previous terms. So u1 plus u2 equals u3, u2 plus u3 equals u4, and so on.
We can generalise this as u subscript n plus u subscript n + 1 equals u subscript n + 2, or you can write that the other way around.
Now we need to write where u1 is 1 and u2 is 1.
Lucas says, "I have started writing a Fibonacci sequence.
It could be denoted this way." Can we write the first five terms of Lucas's sequence? Well done if you spotted, but we've not got enough information.
To generate a Fibonacci sequence, we need to know two numbers.
Lucas says, "Okay, u1 is 4 and u2 is 6." Write the first five terms of Lucas's sequence now.
So we've got 4, 6, and add those, we get 10, add those, get 16, add those, get 26.
Jun says, "I have started writing a different sequence.
It can be denoted this way." With u1 equals 2 and u4 equals 12.
Lucas says, "There's no way to know Jun's sequence as we could not work out the terms." Do you agree with Lucas? Well, let's have a look.
Let's write the rule for u4 first.
So u4 is gonna be u2 add u3, u3 would be u1 add u2.
Well, now we can substitute that in.
If we've got a way of writing u3, we can write u4 as u2 plus u1 plus u2.
Let's substitute the values we know.
So 12 would equal u2 plus 2 plus u2 again.
So 12 would be 2 lots of u2 plus 2.
We can solve that to get that u2 must be 5.
So if u2 is 5, then we get the sequence 2, 5, 7, 12.
You can use a letter to represent the missing u2 if you find that easier.
You may have done that before.
If you make the next term A, then you can write an expression for the third term and then the fourth term and form equation.
A similar method can be used to work out the term-to-term rule given two terms in a sequence.
So here's an example.
A sequence has formed u subscript n + 1 equals k lots of un minus 5, where u1 equals 4.
Given that u3 equals 16, what must the value of k be? So we need to use the information that we have.
So we know what u1 is, we know to get u2, we do k lots of u1 minus 5.
So if we substitute in u1, we get k lots of 4 minus 5 or 4k minus 5.
That's an expression now for the second term.
Well, to get u3, we do k lots of u2 minus 5.
So what we can do now is substitute our expression for u2.
So u3 would be k lots of 4k minus 5 minus 5.
Expand the brackets, we get 4k squared minus 5k minus 5.
And we know u3 is 16, we've got 16 equals 4k squared minus 5k minus 5.
Then all we need to do is rearrange and we have a quadratic to solve.
So set it equal to zero and factorised.
We have two solutions, k is -7/4 or k is 3.
Now we can check our answers.
We do 4 multiply by 3 minus 5 is 7, multiply by 3 minus 5 is 16, and that was our third term.
Let's try it with our other value for k, multiply by -7/4 subtract 5, -12, multiply by -7/4 subtract 5 again, and we get 16 which is our third term.
That is definitely one of the trickiest things that you can do using this notation at the moment.
So well done if you followed along with that.
As always, when you are looking at something new, it can look more complicated than it is.
Right, true or false.
You can generate a Fibonacci sequence as long as you know any two terms. What do you think? Yeah, that's true.
You can form an equation to find the missing terms based on the two known terms. It is easier if they're consecutive.
Right, which of those show the Fibonacci sequence with first five terms, 8, -5, 3, -2, 1.
What do you think? Yeah, it is C.
We're just getting used to that notation.
We should have u subscript n + 2 equals u subscript n plus u subscript n + 1 where u1 is 8 and u2 is -5.
Time for practise.
I'd like you to use these forms to work out the required term.
Give this a go.
This time, I'd like you to generate the first five terms in each of these sequences.
Look carefully at the relationships and at the two terms that you know.
Right, Lucas is trying to find the first five terms at the sequence below.
This is a Fibonacci sequence where we know the first term and then the fourth term.
Most of his method is visible.
I'd like you to fill in the gaps and then write the first five terms of the sequence.
All right, couple of trickier problems for you to solve now.
These are similar to the ones that we did in the example, so you use your notes to help you.
Give these a go.
And your last challenge similar to the last question just got to apply a few more of your algebra manipulation skills.
See how you go.
Well done, I'm really pleased with how well you've engaged with this new way of writing sequences.
So the 5th term in the sequence is gonna be 20.
For B, it's 83.
For C, notice we're given the 8th term, so actually it doesn't take too long to find the 10th term.
That's -1.
For D, we've gotta work backwards from u3 to get u1 is 3.
And E, again, we're working backwards, the first term is 4.
For 2A, we should get the sequence 5, 2, 7, 9, 16.
For B, -3, 4, 1, 5, 6.
Notice for C, we're just swapping the first two terms over.
So we've got 4, -3, 1, but they're -2, -1, so it does matter which way around your first two terms are in a Fibonacci sequence.
And then D, well done if you spotted that we were given the second and the fourth term, so we should have 4, 8, 12, 20, 32.
Right, and then for Lucas's method.
We should have u4 equals u2 plus u3, u3 equals u1 plus u2.
Then substituting in, we've got u4 equals u2 plus u1 plus u2.
Now it looks like he's gonna substitute known values.
We've got 20 equals u2 plus 2 plus u2, so 20 equals 2 lots of u2 plus 2.
And so that equation we get u2 as 9.
Right, we're gonna have 2, 9, 11, 20, 31.
Then let's have a look at our challenges then.
So this is exactly the same as what Lucas did.
You're just having a go yourself.
You end up with the equation 15 equals 2 lots of u2 plus 1, so u2 is 7.
Your sequence is 1, 7, 8, 15, 23.
Then for five, we need to write some expressions.
So u2 can be written as k lots of u1 plus 1.
We substitute in u1 as 1, we get u2 equals k plus 1.
Let's look at u3.
So u3 can be written as k lots of u2 plus 1, u3 then is then k lots of k plus 1 plus 1.
We can write an equation.
So k squared plus k plus 1 equals 31.
Set equals to zero and factorised, you get k for 6 or k minus 5 equals 0.
So k is -6 or k is 5 and you could check that works as well.
Your final challenge.
Exactly the same method as before, you just end up with a trickier quadratic.
So we've got 6k squared minus 7k minus 33 equals 0.
Then you can use whichever method for solving quadratics you like.
I factorised.
So you get 6k plus 11 and k minus 3 equals 0.
So k is -11/6 or k equals 3.
And again, you could check that works.
If you found those last three challenges a little bit tricky, don't panic.
There's two things going on here.
We're learning a new notation and we're applying it to problems. So it's perfectly acceptable if you're finding some of those things a little bit tricky.
Hopefully, what you've seen today is this new way of writing these sequences it's really good for writing a term-to-term rule.
And hopefully, you're also getting used to using this subscript that may well come up in other areas of your mathematics studies.
Well done.
If you want a quick reminder of some of those key pieces of notation we've learned today, please pause the video and read through that now.
Thank you for persevering with that new idea.
As with everything that's new, takes a little bit longer to become an expert.
So feel free to spend a bit more time just going back through some of those questions and making sure you're happy.
Thank you for working with me today and I really look forward to you joining us again for another lesson.