# Reasonably simple series question

1. Dec 14, 2009

### henryc09

1. The problem statement, all variables and given/known data
If you invest £1000 on the first day of each year and interest is paid at 5% on your balance at the end of each year how much money do you have after 25 years?

2. Relevant equations

3. The attempt at a solution
Been a while since I did series, and I was never very good with them. I'm guessing you have some sort of combination of an arithmetic and geometric series, just not really sure how to set it up. Any points in the right direction would be appreciated

2. Dec 14, 2009

### Mentallic

Lets take a look at each year separately, I'll denote the nth year as Pn.

$P_0=1000$ : since you invested £1000 at the beginning of the year.

$P_1=1000+1000(0.05)$ : you already have 1000 in the bank, but now you get 5% interest on that money, so you need to add 0.05x of the investment you have to the bank.

$P_2=P_1(0.05)+P_1$ : now notice that we can factorize P2

$P_2=P_1(1+0.05)=P_1(1.05)$ : and in the same way we can factorize P1.

$P_1=P_0(0.05)+P_0=P_0(1.05)$

So now, $P_2=P_0(1.05)^2$ and following the same pattern, $P_3=P_0(1.05)^3$ etc.

So after 25 years?

3. Dec 15, 2009

### henryc09

I don't think this accounts for the fact that you add another £1000 at the beginning of each year though

4. Dec 15, 2009

### HallsofIvy

Look at the amount added each year separately.
The first years £1000 becomes 1000+ 1000(.05)+ 1000(.052)+ ...+ 1000(.0525) which is a geometric series.

The second years £1000 becomes 1000+ 1000(.05)+ 1000(.052)+ ...+ 1000(.0524) which is also a geometric series, etc.

The last years contribution will be just 1000+ 1000(.05).

Each of those can be calculsted by the formula:
$$\sum_{i=0}^n ar^i= a\frac{1- r^{n-1}}{1- r}$$

For the first year's £1000, that is
$$1000\frac{1- .05^{24}}{1- .05}= \frac{1000}{.95}(1- .05^{24})= (1052.63)(1- .05^{24})$$

For the second year's £, that is
$$1000\frac{1- .05^{23}}{1- .05}= \frac{1000}{.95}(1- .05^{23})= (1052.63)(1- .05^{23})$$
etc.

Adding all those together, we have $$(1052.63)(1- .05^{24}+ 1- .05^{23}+ ....+ 1- .05+ 1)= (1052.63)(25- (.05+ .05^2+ ...+ .05^{24})$$
which gives a second geometric series.

5. Dec 15, 2009

### Mentallic

Oh sorry, I misread the question.

6. Dec 16, 2009

### henryc09

I really don't think that works because the answer you get is far too small, I worked it out manually with a calculator and the answer is near £50,000

7. Dec 16, 2009

### Mentallic

Ok I'll do the question again, after reading all the information this time

$$P_0=1000$$

$$P_1=P_0.I+1000=1000(1.05)+1000$$ : I denotes interest. You multiply by 1.05 rather than 0.05 because the interest of 50 is added to the 1000, so $1000(0.05)+1000=1000(1+0.05)=1000(1.05)$

$$P_2=P_1.I+1000=\left(1000(1.05)+1000\right)(1.05)+1000$$

Now, expand $P_2$ and see if you can spot the pattern. Try finding what $P_3$ is to be sure you're getting the pattern right. So what would be $P_25$?

Now, do you know how to tackle the geometric series?

8. Dec 17, 2009

### gabrielh

I do believe, if I understand your question correctly, that the easiest method to solve this problem is by the use of this formula:

$$A = P \left(1 + \frac{r}{n}\right) ^{rt} \large$$

Where:
P is the inital amount/investment
r is the interest rate as a decimal
n is the number of times the amount is compounded per year
t is the number of years
A is the amount after t years

Hopefully that helps.