by Dr. Carol JVF Burns (website creator)
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Simple interest is interest on principal only.
Compound interest is interest that is calculated on the principal amount together with accumulated interest.

Suppose you invest $\,\$1000\,$ at $\,5\%\,$ simple annual interest.
One year later, you'll have $\,\$1000 + (5\%)(\$1000) = \$1050\,$.

Wouldn't it be better to add in $\,\frac{1}{12}\,$ 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 $\,\frac{1}{365}\,$ 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 $\,\$1000\,$ with $\,5\%\,$ interest,
and see how much benefit is gained from:

$\,\$1000\,$ initial deposit, $\,5\%\,$ interest rate
(all units are dollars, rounded to the nearest cent)
Interest is added in at the end of each compounding period.

time compounding annually compounding monthly compounding weekly compounding daily
$1\,$ day $1000$ $1000$ $1000$ $1000+\frac{0.05}{365}(1000)=1000.14$
$2\,$ days $1000$ $1000$ $1000$ $1000.14+\frac{0.05}{365}(1000.14)=1000.28$
$1\,$ week $1000$ $1000$ $1000+\frac{0.05}{52}(1000)=1000.96$ $1000.96$
$2\,$ weeks $1000$ $1000$ $1000.96+\frac{0.05}{52}(1000.96)=1001.92$ $1001.92$
$1\,$ month $1000$ $1000+\frac{0.05}{12}(1000)=1004.17$ $1004.17$ $1004.18$
$2\,$ months $1000$ $1004.17+\frac{0.05}{12}(1004.17)=1008.35$ $1008.36$ $1008.37$
$6\,$ 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 $\,\$1648.66 - \$1628.89 = \$19.77\,$ by compounding daily versus annually.
As the length of time and the amount of money invested increase, though, the savings do go up—and every little bit helps!


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 $\cssId{s77}{P+\frac{r}{n}\cdot P} \cssId{s78}{=P(1+\frac{r}{n})} \cssId{s79}{=P{(1+\frac{r}{n})}^1}$ factor out $\,P\,$
$2\,$ compounding periods $\cssId{s82}{P(1+\frac{r}{n})+\frac{r}{n}\cdot P(1+\frac{r}{n})} \cssId{s83}{=P(1+\frac{r}{n})(1+\frac{r}{n})} \cssId{s84}{=P{(1+\frac{r}{n})}^2}$ factor out $\,P(1+\frac{r}{n})$
$3\,$ compounding periods $\cssId{s87}{P{(1+\frac{r}{n})}^2+\frac{r}{n}\cdot P{(1+\frac{r}{n})}^2} \cssId{s88}{=P{(1+\frac{r}{n})}^2(1+\frac{r}{n})} \cssId{s89}{=P{(1+\frac{r}{n})}^3}$ factor out $\,P{(1+\frac{r}{n})}^2$
$n \text{ compounding periods} = 1 \text{ year}$ $P{(1+\frac{r}{n})}^n$ notice the emerging pattern
$2n \text{ compounding periods} = 2 \text{ years}$ $P{(1+\frac{r}{n})}^{2n}$
$tn \text{ compounding periods} = t \text{ years}$ $\cssId{s98}{A} \cssId{s99}{=P{(1+\frac{r}{n})}^{tn}} \cssId{s100}{=P{(1+\frac{r}{n})}^{nt}}$ the compound interest formula!

Thus, we have:

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 year).
The amount, $\,A\,$ (principal plus interest), that you have after $\,t\,$ years is given by
the compound interest formula: $$\cssId{s112}{A=P{(1+\frac{r}{n})}^{nt}}$$
Suppose $\,\$2500\,$ is invested at $\,3\%\,$ annual interest, compounded daily, for seven years.
How much money will you have?
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 $(3\%)(\$2500) = \$75\,$.
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:
$P=2500\,$, $\,r=0.03\,$, $\,n=365\,$, $\,t=7\,$
    $\displaystyle \cssId{s131}{A} \cssId{s132}{=2500{\left(1+\frac{0.03}{365}\right)}^{(365\cdot 7)}} \cssId{s133}{=\ \$3084.17}$
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.
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(MAX is 20; there are 20 different problem types.)