# Compound interest + New capital each year

• Mar 15th 2011, 10:01 PM
Elvyne963
Compound interest + New capital each year
Hi, I'm not sure it this formula needs calculus, but I think so. So here is the problem.

You want to invest some money. Let's say the first year you invest 1000$, and you will always reinvest the money you gained, plus every year you will add a new 1000$. You make an interest of 10% each year.

Here is what I have been able to do up to now

x' = ( x + 1000 ) * 1.1 ex : after the first ( 0 + 1000 ) * 1.1 = 1100
x'' = ( x' + 1000 ) * 1.1 second year ( 1100 + 1000 ) * 1.1 = 2310
x ''' = ( x'' + 1000) * 1.1 and so on and so on.

How can I put it in one formula that I can compute and only have to enter the number of year I invested it ?

You get a general formula by looking at what you are doing in general. The first year you invested "A" and at the end of the year had 1.1A. You added another "A" to that to make 1.1A+ A and, at the end of that year had $1.1(1.1A+ A)= 1.1^2A+ 1.1A$. You added another "A" and so had $1.1^2A+ 1.1A+ A$ and, at the end of that year had $1.1(1.1^2A+ 1.1A+ A)= 1.1^3A+ 1.1^2A+ 1.1A$. Now, it should be easy to see that, at the beginning of the nth year, you have $(1.1^n+ 1.1^{n-1}+ \cdot\cdot\cdot\+ 1.1^2+ 1.1+ 1)A$. The part in parentheses is geometric sum, of the form [/tex]a+ ab+ ag^2+ \cdot\cdot\cdot+ ab^{m-1}+ ab^n[/tex] with a= 1 and b= 1.1. It is well known that this sum is equal to $\frac{a(1- b^{n+1}}{1- b}$. Here, that would be $\frac{1- 1.1^n}{1-1.1}= 10(1.1^n- 1)$