Lesson video

Thank you for joining me for today's lesson.

My name is Miss 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 advanced problem solving with further sequences.

We're gonna use all our knowledge about sequences to solve some problems today.

We're gonna talk about interest.

Interest is money added to savings or loans.

Compound interest is calculated on the original amount and the interest accumulated over the previous period.

We're gonna talk a little bit about exponential equations.

The general form is y equals ab to the power of x, where a is the coefficient, b is the base, and x is the exponent.

So let's start by looking at compound interest.

Izzy is investigating different types of loans.

She's used an online loan calculator for borrowing 80 pounds from a loans company.

If no repayments are made, this is how much you would owe each year.

How much is the loan increasing by each year? Do a quick check.

Right, we've got 40, 60, 90, then 135.

Why is this happening? Can you explain what's going on here? Right, well, the loan is increasing because loans gather interest the longer they are left unpaid.

If this is compound interest, then the interest will increase over time.

If we write these values as a sequence, we can explore this further.

This is not a linear sequence, as we can see that the interest is increasing each year.

It's also not a quadratic sequence 'cause there's not a linear pattern in the differences.

I don't know if you saw that before.

What other type of sequence could it be and how could we check? You may have suggested geometric.

If you did, we'd check by looking for a common ratio.

120 divided by 80 is 1.

5, 180 divided by 120 is 1.

5, and 270 divided by 180 is 1.

5.

Looks like a geometric sequence so far.

If a loan is being charged with compound interest, then the value of the loan over time will form a geometric sequence.

We're assuming no repayments are made.

So this is the one we had before.

The common ratio is 1.

5.

What does that mean about the interest rate? Yeah, 1.

5 is the multiplier to increase by 50%.

This loan is gathering interest then at a rate of 50% compounded every year.

And we know we can use multipliers to apply compound interest, so that tells us that this is gonna form geometric sequences.

Izzy says, "This was a payday loans company, which is probably why the yearly interest rate was so high." Yeah, 50% interest rate a year seems really high.

If you're borrowing only for a few days, the yearly equivalent interest rate is gonna be really high 'cause they're never expecting you to borrow the money for that long.

If you're borrowing money for something like a mortgage, which is gonna be over 25 years potentially, then the interest rates are a lot smaller than that per year.

Something you might be interested in exploring.

We can use what we know about calculating compound interest to find nth term rules for this sequence.

So year one is 80 times 1.

5, which is 120.

So notice we're starting at the second number in the sequence that we've written there.

Year two is 80 times 1.

5 squared, year three is 80 times 1.

5 cubed, and year four would be 80 times 1.

5 to the power of four.

How could we generate the nth term then? 80 times 1.

5 to the power of n.

That's assuming we're starting on the number, 120, as our first term.

The general form for an exponential equation is y equals ab to the power of x, where a is the coefficient, b is the base, and x is the exponent.

The nth term of a geometric sequence will be an exponential expression.

Using this knowledge can help us efficiently solve problems with compound interest.

Should we look at this previous example? 80 is our loan amount, 1.

5 is our interest rate as a multiplier, and that's the nth term for the sequence starting on year one.

Right, another loans company charges an interest rate of 10% per year.

If 80 pound was borrowed, what would the first three years look like if no repayments were made? Right, we've got 80 times 1.

1, 80 times 1.

1 squared, and 80 times 1.

1 cubed.

What would the nth term rule for this sequence be? We're gonna have 80 times 1.

1 to the power of n.

80 being our loan amount and 1.

1 being the interest rate as a multiplier.

Right, quick check.

50 pound is borrowed from a loans company charging 2% yearly compound interest.

Which of these would be the nth term rule for the sequence formed by the yearly loan values? What do you think? Well done if you said 50 times 1.

02 to the power of n.

To increase by 2%, we can multiply by 1.

02, and we're starting with 50 pounds borrowed.

50 pound is borrowed from a loans company charging yearly compound interest.

The sequence below is the loan value each year for the first four years.

Which of these is the nth term rule for this sequence? Well done if you picked a.

We can find the multiplier by dividing successive terms and we get 1.

4.

So it's 50 times 1.

4 to the power of n.

A loan company charges 2% compound interest a month.

If 1,000 pounds was borrowed, how much will be owed after five months? Well, we could start by working out the nth term rule.

So that would be 1,000 times 1.

02 to the power of n.

If you want the fifth term, we can substitute n as five.

You'd have done this before when you've done compound interest calculations.

You can always draw a table to help you if you need to.

Now someone else borrowing from this company owes 663 pounds after one month.

How could we work out how much was borrowed? Right, we can't write a complete nth term 'cause we don't know the loan amount.

So I'm gonna write a times 1.

02 to the power of n.

Now I need to find out a.

We can form an equation from the values we know.

So I know that a ties 1.

02 to the power of one is gonna be 663.

To work out a then, I just need to do 663 divided by 1.

02, which is 650.

The loan amount was 650.

You've probably done something similar before with compound interest.

We're just setting up equations now using what we know about these nth term rules.

The nth term rule would be 650 times 1.

02 to the power of n.

Quick check then.

The amount owed on a loan is 2,000 pounds after five years and 2,600 pounds after six years.

What's the interest rate as a percentage? Well done if you said 30%.

We divide our terms. We get a multiplier of 1.

3.

That's a 30% interest rate.

Some of our sequences skills will help us solve trickier problems. For example, after two years, a loan is worth 1,000 pounds, but after four years, this rises to 1,210.

Let's see if we can work out the interest rate on this loan.

We don't know the amount after one year or after three years, so I've drawn some gaps.

I've got to multiply by the common ratio and then multiply by the common ratio again to get from 1,000 to 1,210.

So if we divide the terms, we get the overall multiplier.

That's 1.

21.

So r squared must be equal to 1.

21.

How are we gonna find r? Of course, we need to square root, and that gives us 1.

1.

The interest rate then is 10%.

What is the nth term rule for this sequence? Well, we have the interest rate.

We just need the start value.

We know after two years, we have 1,000 pounds.

So we do 1,000 divided by 1.

1 squared.

We get 826.

826 times 1.

1 to the power of n.

Right, what calculation would find the common ratio for this geometric sequence? What do you think? Right, we need the fourth root of the overall multiplier.

Then divide those terms. Find the fourth root.

The fourth root is actually 1.

5.

A loan of 300 pounds is taken out with monthly compound interest of 3%.

How many months will it take for the loan to go over 500 pounds? Andeep says, "We can start by working out the nth term rule." So it's 300 times 1.

03 to the power of n.

We need to know when this expression is greater than 500.

What could we do? We could keep trying values of n until we find the correct one.

If I try n is five, I get roughly 347.

There's n is 10.

There's n is 15.

There's n is 20.

So I can see it's somewhere between 15 and 20.

So let's look at 16, 17, 18.

I can stop there.

I can see that it's between 17 and 18 months.

As the loan increased monthly, we know it's gonna be after 18 months.

Andeep says, "Because we know the nth term rule, we could plot this on a graph instead." So if you wanted to, you could use the table function on a graph and type in the nth term rule.

It generates the sequence for you.

So we've got y equals 300 times 1.

03 to the power of x.

And y equals 500 is our straight line to help us see.

And we can see the first point with a y value over 500 is when x is 18.

Time to put this into practise.

6,250 pounds is borrowed at a compound interest rate of 4% each month.

See if you can answer those four questions.

A loan model for borrowing money to buy a washing machine is given below.

Andeep says, "I don't think this loan charges compound interest." Is he correct? Write a sentence to explain your answer.

Question three, a loan gathers compound interest yearly.

If no repayments are made, 16,820 pounds will be owed after two years and 24,389 after three years.

Can you answer those four questions for that scenario? Question four, 1,000 pounds is borrowed from a company charging yearly compound interest.

After two years, 1,440 pounds was owed.

Can you answer those questions? Question five, 3,000 pounds was borrowed from company.

After three years, 3,993 was owed.

How much will be owed after eight years? And finally, money was borrowed from a company charging yearly compound interest.

After one year, 10,156,250 pounds was owed.

After five years, that's 11,881,376.

What is the interest rate? And how many years will it take for this loan to be over 20 million? Off you go.

Let's look at our answers then.

We should have 6,500, 7,030.

40.

For a year, it's 12 months.

If we use a power of 12, we get 10,006.

45.

The nth term rule will be 6,250 times 1.

04 to the power of n.

For question two, Andeep is correct.

The values seem to be forming a linear sequence, not a geometric sequence.

Three pound is added on each month.

This suggests the interest is the same each month, which would be simple interest, not compound interest.

So the interest rate for this loan, if we divide the successive terms, we get 1.

45, which is 45% per year.

After five years, we get 51,277.

87.

In order to find out what we borrowed to begin with, we need to divide by 1.

45 squared, and we get 8,000 pounds.

The nth term rule then is 8,000 times 1.

45 to the power of n.

For question four, the overall multiplier is 1.

44.

We find the square root.

We get 1.

2, which is 20% compound interest each year.

The nth term rule then would be 1,000 times 1.

2 to the power of n.

How much be owed after five years? We can just substitute n as five.

2,488.

32.

For question five, it's gonna be useful to find the interest rate first.

So we divide those terms and we need the cube root, which gives us 1.

1.

We've got 3,000 times 1.

1 to the power of eight, which is 6,430.

77.

For question six, we need to divide the terms we know, and then we need that exact value, so don't round that at all 'cause we need the fourth root of that, and that gives us 1.

04.

4% compound interest a year.

Now how many years would it take for this to be over 20 million? Well, it'd be helpful to start by finding the original amount.

So we do 10,156,250 and divide it by 1.

04.

We get our loan amount.

So our nth term is 9,765,625 times 1.

04 to the power of n.

Then we just need to use a bit of trial and improvement until we get the right values.

18 does not give us a value over 20 million, but 19 does, so it's gonna take 19 years.

Well done.

Now we've explored sequences with compound interest, we're gonna have a look at sequences with proof.

We're gonna do a quick recap on proof.

When proving a conjecture, it needs to be shown true for all relevant cases.

Often, the easiest way is to generalise and manipulate into the required form.

So we gotta prove that the sum of three consecutive odd numbers is a multiple of three.

We need to start by generalising an odd number.

What is the general form of an odd number? You could have 2n plus one or or 2n minus one, where n is an integer.

And that is really important, otherwise these would not always be odd.

So the first line of our proof will define any variables and explain any generalisations.

Take three consecutive odd numbers.

2n plus one, 2n plus three, and 2n plus five where n is an integer.

Their sum can be written as 2n plus one plus 2n plus three plus 2n plus five.

I can simplify that to 6n plus nine.

How can we show this is a multiple of three? What would you do? We need to factorise.

We can write this as three lots of 2n plus three.

And don't forget your conclusion.

The sum of three consecutive odd integers is a multiple of three.

And that was a quick recap on how we form a proof.

Which of these will always be a multiple of five when n is an integer? Well done if you said d.

We can show this by factorising to get five lots of 3n plus two.

Which of these will always be a square number when n is a positive integer? You can have a or d.

If n is an integer, n plus one is an integer, and then we've got an integer squared.

For d, we can factorise this to n plus four all squared.

So that is an integer squared.

Right, well, let's put this together with our quadratic sequences.

If we know the general form for the nth term of quadratic sequences, we can show or prove certain results with quadratic sequences.

So we're gonna show that the sum of two quadratic sequences will be a quadratic sequence when the coefficients of n squared in the nth term formulas are not a zero pair.

So let's take any quadratic sequence with nth term rule, an squared plus bn plus c where a, b, and c are constants and a is not equal to zero.

Then we'll have another one of the form pn squared plus qn plus r where p, q, and r are constants and p is not equal to zero and a and p are not a zero pair, so a is not equal to negative p.

Right, now we can add them together.

Collecting like terms, we've got a plus p lots of n squared, b plus q lots of n plus c plus r.

This will be a quadratic expression when a plus p is not equal to zero.

Therefore, the sum of two quadratic sequences will be quadratic when their coefficients of n squared are not a zero pair.

We're gonna look at the nth term rule for triangular numbers because you're gonna use this in a task.

So those are our triangular numbers.

We can show this with a visual representation.

So I've picked the third triangular number.

You could take any triangular number.

I can double it, and then with two lots, I can arrange it into a rectangle.

The dimensions are three dots by four dots.

We can do the same for the next triangular number.

Let's try when n is four.

Double it.

Arrange it into a rectangle.

What are the dimensions of this rectangle? We've got four dots by five dots.

If we doubled the fifth triangular number and arranged the dots into a rectangle, what do you think the dimensions are gonna be? Let's try that.

And we get five dots by six dots.

You might have spotted a pattern.

There's the first and the second triangular numbers as well.

So we've got a one by two, a two by three, a three by four, a four by five, and a five by six.

How could we write the dimensions for the nth pattern? Right, n and n plus one.

So the number of dots in this rectangular pattern can be written as n multiplied by n plus one.

But remember, we started by doubling our triangular numbers to make the rectangles, so now we need to half that to represent the triangular numbers.

So the nth term rule for the triangular numbers is n multiplied by n plus one all over two.

Pause the video and try out a couple of terms to see if this works.

Yeah, it does.

If you take the term number, multiply it by the next term number, and then divide by two, you get the triangular number.

And this is quite useful if you want to find out some of the larger triangular numbers.

We could also use the common second difference to find the nth term of a quadratic sequence.

You will have used this method before.

Let's try using it for the triangular numbers.

So find the common second difference, which is one.

Half the common second difference to get the coefficient of n squared.

So the nth term is in the form 1/2n squared plus something.

Now we need to write the sequence of 1/2n squared.

There's n squared and half it.

I'm gonna use decimals at the moment to help me spot what's happening here.

Now we need to compare our sequence to this sequence, so I'll write it underneath.

And we need to add 0.

5, then add one, then add 1.

5, then add two, and so on.

What do you notice? Yeah, we have a linear sequence.

The nth term rule is 1/2n or and 0.

5n.

So that means the nth term rule of the triangular numbers is 1/2n squared plus 1/2n.

Right, we've seen two ways of writing the nth term rule now for triangular numbers.

I'd like you to show that these are equivalent, Right, well, if we expand the numerator, we get n squared plus n all over two.

So that's another form of the triangular numbers you can use.

And then writing as separate fractions, we can write that as 1/2n squared plus 1/2n.

If a triangular number is expressed as n multiplied by n plus one all over two, where n is a positive integer, how would we express the next triangular number? What do you think? And we need to add one to n.

So that's gonna give us n plus one multiplied by n plus one plus one or n plus two all over two.

So it's that second one.

This is gonna be useful for some of our proofs in a moment.

Right, your turn.

I've drawn some very pretty hexagonal number sequences and pentagonal number sequences.

I would like you to find the nth term of these two quadratic sequences.

Wonder how they compare to the triangular numbers and the square numbers.

Right, the first four terms in a linear sequence are five, eight, 13, 18.

I'd like you to find the nth term for this sequence.

Then a new sequence is created by squaring each term and then adding one to each term.

Prove that all terms in the new sequence are divisible by five.

It's a proof, so all variables need to be defined and it needs a conclusion.

Jacob discovers that when any odd number greater than one is squared, then one is subtracted, then that is divided by eight, you get a triangular number.

He wants to prove this always works.

I would like you to fill in the first step of this proof.

Then can you finish off his proof? Use as many steps as you need.

I wonder if you know this.

The sum of any two consecutive triangular numbers is a square number.

Now I'd like you to prove that this always works.

Give it a go.

And finally, the difference between the squares of consecutive triangular numbers is the cube numbers.

So if you square two consecutive triangular numbers and find the difference, you'll get a cube number.

I'd like you to show this for one example of your choosing.

Then I've started the proof off for you.

Can you carry it on? You'll notice that I've written one triangular number as n times n plus one all over two, and then the previous triangular number as n minus one multiplied by n all over two.

This just makes the proof a little bit easier than using n lots of n plus one all over two and n plus one times n plus two all over two.

Right, see if you can get to the end of that proof.

Off you go.

Let's have a look.

So the common second difference is four.

So we've got two n squared.

But to get to our sequence, we have to subtract one, then two, then three, then four.

So the nth term of the hexagonal numbers is two n squared minus n.

For the pentagonal numbers, we've got a common second difference of three, half that to get 1.

5.

1.

5n squared is 1.

56, 13.

5, 24, and so on.

To get to our sequence, we need to subtract 0.

5, subtract one, subtract 1.

5, subtract two.

Our overall nth term then is 1.

5n squared minus 0.

5n, or 3/2n squared minus 1/2n.

If you did write out the nth term of the triangular numbers, the square numbers, the pentagonal numbers, then the hexagonal numbers, you will see a pattern forming.

Question three, the other term of our linear sequence is 5n minus two.

So squaring and adding one to the sequence, 5n minus two, gives 5n minus two all squared plus one.

Now we can do some rearranging.

Product of two binomials.

We've got 25n squared minus 10n minus 10n plus four plus one, which is 25n squared minus 20n plus five.

And I want to show this is divisible by five.

If I factorise out five, get five lots of 5n squared minus 4n plus one.

This is always a multiple of five when n is an integer.

Therefore, every term in this sequence is divisible by five.

Well done for remembering your proof skills.

So for Jacob's proof, any odd number can be written as 2n plus one, where n is a positive integer.

We want this to start at the number three.

This is fine when n is a positive integer.

Two times one is two, add one is three.

We actually want to define n as a positive integer because we're going to look at the nth term of a triangular number, and the nth term starts when n is one.

So this time, we are gonna want the form 2n plus one rather than 2n minus one.

Because if we use 2n minus one when n is a positive integer, the first number we'll be looking at would be one, and we need odd numbers greater than one.

Right, now we need to apply the rules.

So we've got to square, subtract one, and divide by eight.

With a bit of algebraic manipulation, we get to 1/2n squared plus 1/2n.

Now this is the nth term rule of the triangular numbers.

Therefore, any odd number greater than one squared, subtract one, then divide by eight is a triangular number.

Right, and even more interesting is this proof that sum of any two consecutive triangular numbers is a square number.

We can write any triangular number in this form where n is a positive integer.

You might have used other forms. The next triangular number then would be n plus one times n plus two all over two.

If you decide to go for the previous triangular number instead, that's absolutely fine.

Then I need to add these together.

Lots of algebraic manipulation.

And we get n squared plus 2n plus one.

Well, I can factorise that into n plus one all squared.

This is a square number when n is a positive integer.

Therefore, the sum of any two consecutive triangular numbers is a square number.

Well done if you got to the end of that one.

So you could've picked any examples for these.

So for example, three squared minus one squared is eight.

Six squared minus three squared is 27, which is a cube number.

So you could've tried that for any two consecutive triangular numbers.

So let's finish off our proof.

The differences can be written like this.

It's up to you what stages of algebraic manipulation you're gonna do when.

I've decided to expand the brackets within the squares.

Then I've squared the numerators and squared the denominators.

I've done that in a couple of stages 'cause it was tricky.

I've got a common denominator of four, so I can combine my fractions.

Then I've gotta make sure I'm subtracting everything, so I've used brackets to help me.

If you do that, you get 4n cubed over four, which is n cubed, which is the nth term of the cube numbers.

Therefore, the difference between the squares of consecutive triangular numbers is the cube numbers.

Well done for that proof.

There was lots of tricky algebraic manipulation there.

Good way of testing your skills.

This fact can actually be used to show that the only number which is triangular and cube is one.

That's something you might want to look into.

Well done for all your hard work today.

We've spent a lot of time looking at compound interest and geometric sequences, and we're now starting to be able to find the nth term of a geometric sequence.

That's quite helpful.

And then we had lots of fun playing around with proving results with triangular numbers.

Knowing the nth term of a triangular number can be quite helpful.

At least knowing how to find that is useful.

And we looked at two ways of doing that, didn't we? We did it with a dot pattern and doubling the dot pattern and making rectangles, and we did that using a more formal method.

You've worked really hard today.

Well done.

Thank you for joining us, and I look forward to seeing you in a future lesson.