# Introduction, sequences & Series, future value of payments  (Page 5/5)

 Page 5 / 5

Can you see what is happened? Making regular payments of R2 000 instead of the required R1,709,48, you will have saved R76 675,20 (= R410 275,20 - R333 600) in interest, and yet you have only paid an additional amount of R290,52 for 166,8 months, or R48 458,74. You surely know by now that the difference between the additional R48 458,74 that you have paid and the R76 675,20 interest that you have saved is attributable to, yes, you have got it, compound interest!

## Future value of a series of payments

In the same way that when we have a single payment, we can calculate a present value or a future value - we can also do that when we have a series of payments.

In the above section, we had a few payments, and we wanted to know what they are worth now - so we calculated present values. But the other possible situation is that we want to look at the future value of a series of payments.

Maybe you want to save up for a car, which will cost R45 000 - and you would like to buy it in 2 years time. You have a savings account which pays interest of 12% per annum. You need to work out how much to put into your bank account now, and then again each month for 2 years, until you are ready to buy the car.

Can you see the difference between this example and the ones at the start of the chapter where we were only making a single payment into the bank account - whereas now we are making a series of payments into the same account? This is a sinking fund.

So, using our usual notation, let us write out the answer. Make sure you agree how we come up with this. Because we are making monthly payments, everything needs to be in months. So let $A$ be the closing balance you need to buy a car, $P$ is how much you need to pay into the bank account each month, and $i12$ is the monthly interest rate. (Careful - because 12% is the annual interest rate, so we will need to work out later what the monthly interest rate is!)

$A=P{\left(1+i12\right)}^{24}+P{\left(1+i12\right)}^{23}+...+P{\left(1+i12\right)}^{1}$

Here are some important points to remember when deriving this formula:

1. We are calculating future values, so in this example we use ${\left(1+i12\right)}^{n}$ and not ${\left(1+i12\right)}^{-n}$ . Check back to the start of the chapter if this is not obvious to you by now.
2. If you draw a timeline you will see that the time between the first payment and when you buy the car is 24 months, which is why we use 24 in the first exponent.
3. Again, looking at the timeline, you can see that the 24th payment is being made one month before you buy the car - which is why the last exponent is a 1.
4. Always check that you have got the right number of payments in the equation. Check right now that you agree that there are 24 terms in the formula above.

So, now that we have the right starting point, let us simplify this equation:

$\begin{array}{ccc}\hfill A& =& P\left[{\left(1+i12\right)}^{24}+{\left(1+i12\right)}^{23}+...+{\left(1+i12\right)}^{1}\right]\hfill \\ & =& P\left[{X}^{24}+{X}^{23}+...+{X}^{1}\right]\mathrm{using}\mathrm{X}=\left(1+\mathrm{i}12\right)\hfill \end{array}$

Note that this time X has a positive exponent not a negative exponent, because we are doing future values. This is not a rule you have to memorise - you can see from the equation what the obvious choice of X should be.

Let us re-order the terms:

$A=P\left[{X}^{1}+{X}^{2}+...+{X}^{24}\right]=P·X\left[1+X+{X}^{2}+...+{X}^{23}\right]$

This is just another sum of a geometric sequence, which as you know can be simplified as:

$\begin{array}{ccc}\hfill A& =& P·X\left[{X}^{n}-1\right]/\left(\left(1+i12\right)-1\right)\hfill \\ & =& P·X\left[{X}^{n}-1\right]/i12\hfill \end{array}$

So if we want to use our numbers, we know that $A$ = R45 000, $n$ =24 (because we are looking at monthly payments, so there are 24 months involved) and $i=12%$ per annum.

BUT (and it is a big but) we need a monthly interest rate. Do not forget that the trick is to keep the time periods and the interest rates in the same units - so if we have monthly payments, make sure you use a monthly interest rate! Using the formula from Grade 11, we know that $\left(1+i\right)={\left(1+i12\right)}^{12}$ . So we can show that $i12=0,0094888=0,94888%$ .

Therefore,

$\begin{array}{ccc}\hfill 45\phantom{\rule{3.33333pt}{0ex}}000& =& P\left(1,0094888\right)\left[{\left(1,0094888\right)}^{24}-1\right]/0,0094888\hfill \\ \hfill P& =& 1662,67\hfill \end{array}$

This means you need to invest R166 267 each month into that bank account to be able to pay for your car in 2 years time.

## Exercises - present and future values

1. You have taken out a mortgage bond for R875 000 to buy a flat. The bond is for 30 years and the interest rate is 12% per annum payable monthly.
1. What is the monthly repayment on the bond?
2. How much interest will be paid in total over the 30 years?
2. How much money must be invested now to obtain regular annuity payments of R 5 500 per month for five years ? The money is invested at 11,1% p.a., compounded monthly. (Answer to the nearest hundred rand)

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There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
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scanning tunneling microscope
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biomolecules are e building blocks of every organics and inorganic materials.
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how did you get the value of 2000N.What calculations are needed to arrive at it
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