COMPOUND INTEREST FORMULA

If you put $\,\$1000\,$ in a bank that offers $\,5\%\,$ simple annual interest,
then your interest is added in only after one year has passed.
If you withdraw your savings one day before the year is over,
you'll have only your original $\,\$1000\,$ (drat).
However, if you wait the extra day (making one full year),
then you'll have [beautiful math coming... please be patient] $\,\$1000 + (5\%)(\$1000) = \$1050\,$.

Wouldn't it be better to add in [beautiful math coming... please be patient] $\,\frac{1}{12}\,$ of your annual interest after the first month,
giving a slightly greater amount to earn interest for the next month?
Or, why not add in [beautiful math coming... please be patient] $\,\frac{1}{365}\,$ of your annual interest after the first day,
giving a slightly greater amount to earn interest for the next day?

When you add in interest at regular intervals, this is called compound interest.
With compound interest, you are earning interest on your interest, not only on your original principal!
The Compound Interest Formula is a non-recursive formula that is convenient for working with compound interest situations,
and is the subject of this section.

Before deriving the compound interest formula, let's go back to that [beautiful math coming... please be patient] $\,\$1000\,$ with $\,5\%\,$ interest,
and see how much benefit is gained from:

EFFECTS OF COMPOUNDING
$\,\$1000\,$ initial deposit, $\,5\%\,$ interest rate
(all units are dollars, rounded to the nearest cent)

time simple annual interest compounding monthly compounding weekly compounding daily
$1\,$ day $1000$ $1000$ $1000$ [beautiful math coming... please be patient] $1000+\frac{0.05}{365}(1000)=1000.14$
$2\,$ days $1000$ $1000$ $1000$ [beautiful math coming... please be patient] $1000.14+\frac{0.05}{365}(1000.14)=1000.28$
$1\,$ week $1000$ $1000$ [beautiful math coming... please be patient] $1000+\frac{0.05}{52}(1000)=1000.96$ $1000.96$
$2\,$ weeks $1000$ $1000$ [beautiful math coming... please be patient] $1000.96+\frac{0.05}{52}(1000.96)=1001.92$ $1001.92$
$1\,$ month $1000$ [beautiful math coming... please be patient] $1000+\frac{0.05}{12}(1000)=1004.17$ $1004.17$ $1004.18$
$2\,$ months $1000$ [beautiful math coming... please be patient] $1004.17+\frac{0.05}{12}(1004.17)=1008.35$ $1008.36$ $1008.37$
$6\,$ months $1000$ $1025.26$ $1025.30$ $1025.31$
$1\,$ year $1000 + 0.05(1000) = 1050.00$ $1051.16$ $1051.25$ $1051.27$
$2\,$ years $1050 + 0.05(1050) = 1102.50$ $1104.94$ $1105.12$ $1105.16$
$10\,$ years $1628.89$ $1647.01$ $1648.33$ $1648.66$

The added savings from earning interest on interest is perhaps not quite as much as you'd hope.
For example, in one year you'd earn an additional $\,\$1051.27 - \$1050 = \$1.27\,$ over simple annual interest, by adding in interest daily.
In ten years, you'd earn an additional $\,\$1648.66 - \$1628.89 = \$19.77\,$ over simple annual interest, by adding in interest daily.
As the length of time and the amount of money invested increase, though, the savings do go up—and every little bit helps!

DERIVATION OF THE COMPOUND INTEREST FORMULA

The compound interest formula results from using variables to represent a general investing situation,
writing down several computations, and seeing a pattern emerge.

In a nutshell, you're going to invest $\,P\,$ dollars at annual interest rate $\,r\,$,
add in interest $\,n\,$ times per year, and see how much you have after $\,t\,$ years.

Here are the details:

The table below shows the accumulations after various numbers of compounding periods:

after this time... you'll have this much money... NOTE:
$1\,$ compounding period [beautiful math coming... please be patient] $P+\frac{r}{n}\cdot P=P(1+\frac{r}{n})=P{(1+\frac{r}{n})}^1$ factor out $\,P\,$
$2\,$ compounding periods [beautiful math coming... please be patient] $P(1+\frac{r}{n})+\frac{r}{n}\cdot P(1+\frac{r}{n})=P(1+\frac{r}{n})(1+\frac{r}{n}) =P{(1+\frac{r}{n})}^2$ factor out $\,P(1+\frac{r}{n})$
$3\,$ compounding periods [beautiful math coming... please be patient] $P{(1+\frac{r}{n})}^2+\frac{r}{n}\cdot P{(1+\frac{r}{n})}^2 =P{(1+\frac{r}{n})}^2(1+\frac{r}{n}) =P{(1+\frac{r}{n})}^3$ factor out $\,P{(1+\frac{r}{n})}^2$
.........
$n \text{ compounding periods} = 1 \text{ year}$ [beautiful math coming... please be patient] $P{(1+\frac{r}{n})}^n$ notice the emerging pattern
$2n \text{ compounding periods} = 2 \text{ years}$ [beautiful math coming... please be patient] $P{(1+\frac{r}{n})}^{2n}$
$tn \text{ compounding periods} = t \text{ years}$ [beautiful math coming... please be patient] $A=P{(1+\frac{r}{n})}^{tn}=P{(1+\frac{r}{n})}^{nt}$ the compound interest formula!

Thus, we have:

THE COMPOUND INTEREST FORMULA
Suppose you invest $\,P\,$ dollars at (simple) annual interest rate $\,r\,$,
and add in interest $\,n\,$ times per year (that is, there are $\,n\,$ compounding periods per years).
The amount, $\,A\,$ (principal plus interest), that you have after $\,t\,$ years is given by
the compound interest formula: [beautiful math coming... please be patient] $$A=P{(1+\frac{r}{n})}^{nt}$$
USING THE COMPOUND INTEREST FORMULA
Question:
Suppose $\,\$2500\,$ is invested at $\,3\%\,$ annual interest, compounded daily, for seven years.
How much money will you have?
Solution:
It's a good idea to do a crude estimate first to get a ‘ballpark’ figure.
This catches a lot of calculator mistakes.

If there were no compounding at all, then in one year the interest would be [beautiful math coming... please be patient] $(3\%)(\$2500) = \$75\,$.
Do this for $\,7\,$ years:   $7(\$75) = \$525\,$
Put together the principal plus the interest:   $\,\$2500 + \$525 = \$3025\,$
You'll do better than this by adding in interest at regular intervals,
so you know you'll have more than $\,\$3025\,$.

Now, use the compound interest formula:
[beautiful math coming... please be patient] $P=2500\,$, $\,r=0.03\,$, $\,n=365\,$, $\,t=7\,$
    [beautiful math coming... please be patient] $\displaystyle A=2500{\left(1+\frac{0.03}{365}\right)}^{(365\cdot 7)}=\ \$3084.17$
Compare with $\,\$3025\,$; believable!

If you want, you can do the computation up at WolframAlpha (cut-and-paste):

2500(1 + 0.03/365)^(365*7)

(Notice that the exponent computation must be put inside parentheses, to get the correct order of operations.)
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Graphs of Functions


On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
For a simpler presentation, units are suppressed in the calculations, and shown only in the final answer.
(MAX is 21; there are 21 different problem types.)