Reasonably simple series question

[henryc09]

Homework Statement

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?

Homework Equations

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

 

[Mentallic]

Lets take a look at each year separately, I'll denote the n^th year as P_n.

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?

 

[henryc09]

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

 

[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(.05<sup>24) 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.</sup>

 

[Mentallic]

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

Oh sorry, I misread the question.

 

[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

 

[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?

 

[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.