Simple interest is interest on principal only.
Compound interest is interest that is calculated on the principal amount together with accumulated interest.
[To simplify the discussion on this page,
assume that for simple annual interest, no interest is added in until a year has passed.
Similarly, for monthly interest, assume that no interest is added in until a month has passed.
You get the idea!]
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 nonrecursive 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:
time  compounding annually  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 $\,\$1648.66  \$1628.89 = \$19.77\,$ 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  [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:
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. 
PROBLEM TYPES:
