2 Compound Interest
With simple interest, we were assuming that we pocketed the interest when we received it. In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on that interest in future years. This reinvestment of interest is called compounding.
Suppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow?
The 3% interest is an annual percentage rate (APR) – the total interest to be paid during the year. Since interest is being paid monthly, each month, we will earn [latex]\frac{3%}{12}[/latex]= 0.25% per month.
In the first month,
P0 = $1000
r = 0.0025 (0.25%)
I = $1000 (0.0025) = $2.50
A = $1000 + $2.50 = $1002.50
In the first month, we will earn $2.50 in interest, raising our account balance to $1002.50.
In the second month,
P0 = $1002.50
I = $1002.50 (0.0025) = $2.51 (rounded)
A = $1002.50 + $2.51 = $1005.01
Notice that in the second month we earned more interest than we did in the first month. This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the $2.50 of interest we earned the first month. This is the key advantage that compounding of interest gives us.
Calculating out a few more months:
| Month | Starting balance | Interest earned | Ending Balance |
| 1 | 1000.00 | 2.50 | 1002.50 |
| 2 | 1002.50 | 2.51 | 1005.01 |
| 3 | 1005.01 | 2.51 | 1007.52 |
| 4 | 1007.52 | 2.52 | 1010.04 |
| 5 | 1010.04 | 2.53 | 1012.57 |
| 6 | 1012.57 | 2.53 | 1015.10 |
| 7 | 1015.10 | 2.54 | 1017.64 |
| 8 | 1017.64 | 2.54 | 1020.18 |
| 9 | 1020.18 | 2.55 | 1022.73 |
| 10 | 1022.73 | 2.56 | 1025.29 |
| 11 | 1025.29 | 2.56 | 1027.85 |
| 12 | 1027.85 | 2.57 | 1030.42 |
To find an equation to represent this, if Pm represents the amount of money after m months, then we could write the recursive equation:
P0 = $1000
Pm = (1+0.0025)Pm-1
You probably recognize this as the recursive form of exponential growth. If not, we could go through the steps to build an explicit equation for the growth:
P0 = $1000
P1 = 1.0025P0 = 1.0025 (1000)
P2 = 1.0025P1 = 1.0025 (1.0025 (1000)) = 1.0025 2(1000)
P3 = 1.0025P2 = 1.0025 (1.00252(1000)) = 1.00253(1000)
P4 = 1.0025P3 = 1.0025 (1.00253(1000)) = 1.00254(1000)
Observing a pattern, we could conclude
Pm = (1.0025)m($1000)
Notice that the $1000 in the equation was P0, the starting amount. We found 1.0025 by adding one to the growth rate divided by 12, since we were compounding 12 times per year.
Generalizing our result, we could write
[latex]{{P}_{m}}={{P}_{0}}{{\left(1+\frac{r}{k}\right)}^{m}}[/latex]
In this formula:
m is the number of compounding periods (months in our example)
r is the annual interest rate
k is the number of compounds per year.
While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If N is the number of years, then m = N k. Making this change gives us the sta