Lesson video

Hello and welcome to today's video.

I'm really glad that you have decided to learn with us today.

My name is Ms. Davies and I'm going to be helping you as you work your way through this lesson.

Let's get started.

Welcome to our problem solving lesson with non-linear relationships.

You are going to be using your knowledge of non-linear relationships to solve problems today.

If you're not confident with a geometric sequence or what a triangular number is, just pause the video and read through 'cause we're going to use those in the lesson.

This lesson is split into three parts.

We're going to start by looking at problem solving with geometric relationships.

So the game of chess was invented many, many years ago.

And there's a story behind its invention.

So the story goes that the finished game was presented to a mean King who refused to share his vast stores of rice with the people despite there being a famine.

The king liked the game and offered a prize for whatever reward this person wished for.

What this person said was, "I would like a single grain of rice on the first square of the board." Doesn't sound too much, does it? "Two grains on the second square, four grains on the third square, and then doubling each time." Think about it, four grains of rice doesn't seem that much.

Even eight grains of rice is not going to make a dinner.

The King was startled at such a small price for such a wonderful game and he began to count out the rice.

There we go.

We've got one, two, four.

How many grains will there be on the 64th square? You might want to make a prediction and then we're going to work through it.

Just a glance at that top row, we've got one, two, four, eight, 16, 32, 64, 128, right? We've got to 128, which again does not seem too many grains of rice.

However, that's going to take a while to count out, isn't it? Let's look at the next row.

Wow.

So by the end of that next row, we're up to 32,768 grains of rice, and that third row is over 8 million grains by the 24th square.

Let's look at how many there'd be on the 64th square.

You're going to need calculators for this.

We have a look at a rule.

One is two to the power of zero, two is two to the power of one, three is two squared, four is two cubed.

We're multiplying by two each time, so that exponent is going to going to increase by one each time.

So for the 64th square, we're going to end up with two to the power of 63.

You might want to type that into calculators.

Your calculators have probably given you an answer in something called "standard form" because it's so big.

So that is roughly 9x10 to the power of 18.

So that's 9x10x10x10, 18 times.

That is the size of the number we are looking at, nine then with 18 zeroes.

It's the number nine quintillion.

Obviously, the King was unable to pay his debt.

This story, we don't know if it's true or not, but it's a good example of the power of geometric sequences and it's also the limitations of using this element in modelling.

So we know that things can double and that is a sequence we can follow for a while, but real-life examples may not follow that pattern as the numbers get too big too fast.

Let's have a look at this example.

"There are 2000 penguins living on an island.

Every decade their population doubles." So there we've got 2000, each penguins representing a thousand.

Then after a decade there'll be 4,000 penguins, then 8,000, 16,000, 32,000, 64,000.

I'm sure you can see a problem here.

"The island will eventually run out of space.

The waters will run out of food.

When penguins have to swim further for food, they spend more time in the water with their predators so there'll be a natural limit to the population on the island." Let's have a look at what this looks like when we graph it.

So we are doubling each time on the y-axis and the x-axis is showing each decade.

But what we know is somewhere there'll be a natural limit.

There'll be the amount of penguins that can live comfortably and survive on that island.

So growth cannot be endless.

The line would flatten out somewhere.

Time for you to apply this.

"When the sun comes out in spring, flowers grow very quickly." "The flowers height is following a geometric pattern." It's shown in the table to your left.

What could we assume the flower's height will be next week and what problems do you foresee with this growth model in the future? Spend a bit of time, come back to the answers.

So if we follow this geometric pattern of multiplying by three, the plant will be 27 centimetres tall in four weeks.

Hopefully you said something like, "growth is not usually endless." This model will have a natural limit when the plant reaches its natural height.

If it were to continue, then in eight weeks the flower would be over 20 metres tall, which does not sound realistic.

Time for you to have a practise.

So something else that often forms a geometric sequence is the passing of information.

So in this scenario, the bus is going to be late back from the school trip and only one pupil phones their parents.

That parent then phones five other parents, and each of those parents phone five other parents.

For this example, we're going to assume it takes 10 minutes to make those five phone calls.

How many people would know after 40 minutes? Why might this not be realistic? If it carried on, how many people would know after two hours and why might that not be realistic? Try this out, come back when you're ready for the answers.

Let's have a look then.

So if we multiply by five every 10 minutes, then after 40 minutes there'd be 625 plus the original parent who found out first.

So 626 people would know if the growth continued at the above rate.

Why might this not be realistic? Well, some parents might ring the same people, some parents might not ring anyone.

You might have said it might take longer than 10 minutes to ring five parents.

Depending on the size of the school trip, you might be thinking that 625 sounds like too many parents to be rung in the first place.

Let's have a look after two hours if this was to continue.

So down the left hand side you can see it's multiplying by five.

Then I started to get into rather big numbers, so I thought I might think about my exponents and use my calculator.

So after 10 minutes, there's five people.

Then we're multiplying that by five so that's five squared.

Then we're multiplying that by five, which is five cubed, five to power four, and so on.

For two hours then, it'd be five to the power of 12.

That's the amount of 10 minutes in two hours.

Five to the power of 12, and then I've added one for that original parent gives me that value on my calculator.

That is approximately 200 million people.

Well, we know for a start that no school is that big, so there will not be that many parents to tell.

There is going to be a natural limit to that information passing.

Now we're going to have a look at adding sequences.

We're going to investigate the idea of what happens when we add the values in two different sequences together.

Sophia says, "I think adding two arithmetic sequences will give a different arithmetic sequence." I'm going to show you an example.

So if we add 3, 5, 7, 9, 11 to the sequence, 4, 7, 10, 13, 16, so if we add the first terms together, the second terms together and so on, we get the sequence 7, 12, 17, 22, 27.

That's an arithmetic sequence with a common difference plus seven.

I'd like you to try your own example.

What do you think to Sophia's statement? So Sophia is correct for my example.

I wonder if this will always work.

We're going to try with an increasing and a decreasing sequence.

So 5, 8, 11, 14 and 10, 8, 6, 4, we're going to add those together and they're both going to continue with a linear pattern.

When we add them, we have 15, 16, 17, 18.

So we have an arithmetic sequence with a common difference of one.

I wonder if you started noticing any links between the common differences.

We've only tried a couple of examples, we can't say it always works.

Even if we tried a hundred examples, it would not be proof that this always works.

What we can do is we can use algebra to explore the structure and make a proof that it will definitely always work.

So let's look at the nth term rules.

The nth term rule of the first sequence is 2n+1, and the second one is 3n+1.

If we wanted a number in the sequences added together, you could pick a number for N, you could apply the rule 2n+1, you can apply the rule 3n+1, and then you can add them together.

Adding those together, we get 5n+2.

So the rule for our new sequence should be 5n+2.

And if we have a look at it, that seems to be true, we're increasing by five each time, and seven is two more than the first number in the five times table.

That is the correct nth term rule.

So what we can see is adding two sequences means we can add their nth term rules.

If we have two linear expressions, adding them together will always give another linear expression.

You can prove that using algebra.

Sophia says, "What about if we add two quadratic sequences together?" A quadratic sequence has a linear relationship in the differences.

We can say they have a common second difference.

So for example, a sequence might add two, then four, then six, then eight, so the linear pattern in those differences.

Try an example, see what you think is happening.

Here's my example.

So I've got 2, 6, 12, 20, 30, it has a common second difference of +2.

And then 5, 6, 8, 11, 15 with a common second difference of +1.

When I add those together, I get 7, 12, 20, 31, 45.

So that's increasing by five, then eight, then 11, then 14.

So that seems to be a quadratic sequence, this time with a common second difference of three.

We can explore this algebraically, but not without knowing the nth term rules.

For now, we're just going to try some more examples.

Okay, you're going to try this one then.

The square numbers form a quadratic sequence when written in numerical order.

The triangular numbers do, too.

I'd like you to write the first five terms of the sequence of square numbers in ascending order.

Then do the same for the triangular numbers.

And then I'd like you to add your sequences together using the same rules we did on the previous examples.

What do you notice about your new sequence? Give it a go.

So they're your square numbers and your triangular numbers.

Adding together we should get 2, 7, 15, 26, 40.

We have got a quadratic sequence.

It increases by 5, 8, 11, 14 with a common second difference of +3.

If you think about it, your square numbers have a common second difference of two and your triangular numbers have a common second difference of one.

When we added them, we got a common second difference of three.

Time to put that all together then.

So in each question, the sequences were created by adding the terms of two of the sequences below.

So I've given you sequences A to H and they were added to make the sequences in questions A to I.

I'd like you to work out which two sequences were added in each question.

You might want to spend some time looking at your sequences, seeing if you can spot any linear sequences, any geometric sequences, any that have a common second difference, and then that might help you with A to I.

Well done.

So you can use the examples that you looked at previously and any of your own examples, like, to investigate the following questions: write a sentence about any patterns you have noticed.

Okay, so for A you had C+D, B was A+C, C was E+F, D was D+F, E was A+D, and F was D+H.

G was A+B, H was C+G, and I was E+G.

For question two, I wanted the nth term of sequence A.

Sequence A had an nth term of 7n-2.

Then I wanted you to look at the nth term rules for the sequences that were added to get A, which was C and D.

The nth term rule for C is 5n-1 and for D is 2n-1.

When you add 5n-1 and 2n-1, you get 7n-2, which was our rule for A.

Essentially, what we've looked at is that adding the two nth term rules will give you the nth term rule for the sequence when you've added the values together.

Okay, lots of things you could have discovered when you're exploring these.

So when you add two geometric sequences together, interestingly, you do not get another geometric sequence.

In fact, it's hard to see any arithmetic or geometric pattern in the differences or ratios between terms, which was maybe quite a surprising result.

Two quadratic sequences, often you get another quadratic sequence where the common second difference is equal to the sum of the second differences in the sequences that you have added.

So we saw that before with the square and the triangular numbers.

They had a common second difference of two and one.

The new sequence had a common second difference of three.

However, and I wonder if you spotted this, in question I of question one, if the second differences are additive inverses of each other, you can add two quadratic sequences and get a linear sequence.

So if the differences are increasing by two in one sequence but decreasing by two in another sequence, when you add them together, you can end up with a linear sequence.

Most of the time, though, two quadratic sequences add to another quadratic sequence.

For C, a linear sequence for geometric sequence.

So the resulting sequences were not arithmetic or geometric, but you might have seen a pattern.

There seems to be a geometric relationship in the second differences between terms. And finally, a linear to a quadratic sequence, you will get a quadratic sequence.

If you knew the nth term rules, you could add the nth term rules to get the new nth term, and if you add a quadratic expression to a linear expression, you just get another quadratic expression.

None of that is stuff that you need to memorise, but it's really good that you've spent some time looking at those sequences, spotting patterns in the way things develop.

That will help your mathematical reasoning skills as you improve with your mathematics skills.

The last part of the lesson, we're going to look at something called the Collatz conjecture.

I'd like you to have a look at the sequence below.

Can you spot any patterns? Do you think you know what the next value in the sequence might be? Spend some time on this.

Any patterns that you can spot may be helpful.

Off you go.

Okay, you might have noticed that when a number is even, the next term is half.

So 22 halved is 11, 34 halved is 17.

At the end you've got 40, half it to get 20, half it to get 10.

So you might have guessed that the next term would be five, you would be correct.

What happens if the values are odd? If you haven't worked out yet, pause the video and have another look.

Don't worry if you can't spot it, it's a little bit difficult to spot just from a few terms. So if the value is odd, the next term is three lots of the previous term add one.

So 7x3 is 21, add one is 22, and that gets us back to an even number, doesn't it? 11x3 is 33, add one is 34, we're back to an even number.

The same with 17 and 13.

Check them if you'd like to.

One way we can say this is if we call the term A, the next term would be 3a+1 if it's odd.

So this is actually a famous pattern which leads to a famous conjecture called the Collatz conjecture.

Your sequence can start with any positive integer.

If it's even you half it, if it's odd you multiply by three and add one.

And you continue that pattern until you get down to one.

So I'd like you to work out the next five terms using that rule.

Off you go.

Well done.

If you've got 16 and then eight, four, two, and one.

Eventually the sequence reaches the number one.

If we start with a number seven, like we did in this example, the one is the 17th term in the sequence.

So this Collatz conjecture, named after the mathematician Lother Collatz, says that for any positive integer start value, this sequence will always reach the number one eventually.

However, what's really interesting about this conjecture is no one has actually proven that it will always work.

No one's found a mathematical way to prove that this will work, apart from just keep trying every value.

But, of course, it's really hard to try every value up to infinity because there's infinite values.

Mathematicians are still working on this problem today.

Trillions of values have been tested by a computer programme and no counter examples have yet been found.

But we know that this does not mean the pattern is always going to continue.

What mathematicians are looking for is a mathematical way to prove it'll always work.

We're not going to get to that today, but we can spot some patterns.

So we have seen how the sequence is formed starting at seven.

What are the numbers can we now say also get to the number one that we don't need to try out? Of course, any number in the sequence above.

So we know that this works for 13, 'cause 13 would be 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

What number could we start on so one is the sixth number in the sequence? Okay, we have two options.

We could just look at the sequence above and we know that if we start on five, one is the sixth number in the sequence.

But there's another option that would get you to 16, five gets you to 16, but also 32 would get you to 16.

So we could have 32, 16, 8, 4, 2, 1.

There's two different sequences where one is the sixth term.

Let's have a look at 64.

Why is 64 a relatively simple sequence? Hopefully, you spotted that 64 is two to the power of six.

So if you keep halving 64, if you do it six times, you reach the number one.

What we can do then is we can use this to work backwards from sequence to add an extra term.

So if you think about 64, we know that 64 would go 32, 16, 8, 4, 2, 1.

We can work backwards to find out what the previous number could be.

If we double it, the previous number could be 128, but also the previous number could have been an odd number.

So we need a number that when you multiply it by three and add one, you get 64.

If we do the inverse, we need to subtract one and divide by three.

So 63 divided by three gives us 21.

So we could also have 21 and 64.

Okay, time to check that you are happy with this rule.

So these sequences are formed using the Collatz rule.

I'd like you to work out the next two terms in each sequence.

Give it a go and then we'll check our answers.

So we should half to get three and then we need to triple and add one, which gives us 10.

For B, it's odd, so we need to triple 35 and add one, that gives us 106 and then we can half it.

For C, halving 46 gives us 23.

We don't actually need to work out the next number because it's in the sequence above, 23 takes you to 70.

All right, time for you to explore this.

So the Collatz conjecture says that any sequence of positive integers, which follow this rule, will eventually reach the number one.

The rule's there for you to remind yourself.

I would like you to test this conjecture for the integers one to 10.

So write out all the Collatz sequences first starting with one, then two, and three, up to 10.

Then I want you to think, what would happen if we continued these sequences beyond the number one? So instead of stopping at one, what would happen if you applied the rule to the number one? And then what positive integer less than 20 gives the longest sequence before reaching one.

See if you can do that in the most efficient way.

You shouldn't have to write out all the sequences from one to 20.

Some of them you should be able to tell once you've written the first 10.

Give it a go.

Okay, so, I would like you to tell me how many different sequences with one as the fifth term there are.

You might want to start with one and work backwards.

Do the same, how many different sequences with one as the sixth term, then how many different sequences with one as the 10th term can you find? Give that a go and come back for the answers.

Lovely, I'm not going to read out all these sequences so pause the video and read them if you want to check you've got the right ones.

The longest with numbers less than 10 seem to be nine.

Some of them you didn't have to write out once you've done the first three, so, for example, five and four and eight are all in the sequence starting with three so you already knew what they were going to be.

For B, what would happen if we carried it on? We would get a loop where you get one would then take you to four, then two, then one, and that would take you up to four, then two, then one, and it would loop around.

18 and 19 were the longest, they have a 21st term of one.

How many different sequences with one is the fifth term? We've already seen the one we start on 16 and just half, that is the only one.

There's no other way to make the number eight.

There's no integer value that when you multiply it by three and add one, you get eight.

How many different sequences with one as the sixth term are there? Two options, starting on 32 or starting on five.

I wonder if you found all of the ones with one as the 10th term.

There was six in total.

What you have to be careful when you are working backwards is remember it's only odd numbers that you multiply by three and add one.

If you've got an even number that multiplies three and adds one to get a number, that doesn't matter 'cause when we get an even number, we halve it.

Check those six sequences and see how many you found.

Well done today.

We have looked at how the spotting a sequence makes it possible to predict behaviour.

We've looked at applying mathematical modelling to real-life situations, such as the penguins on the island, but we've seen how that has its limitations.

Then we had fun playing around building sequences by adding them together.

We looked at how the nth term rules could be added together to get the rule for the new sequence.

If you go on to do further sequences in the future, that concept of adding the nth term rules together gives us a really nice way of finding nth terms of trickier sequences.

We've seen how there's problems in mathematics that are ongoing, that mathematicians are working to figure out.

I'm hoping you are thinking that'll be a really interesting thing to spend some time doing in the future, that is what mathematics is all about.