What is Percent Increase?
In order to understand compounding, you need to first understand the percentage increase of a quantity.
If P is a quantity that is increased by a percentage rate r, then the new quantity is P + r P
You need to retain the above: A quantity P increased by a percentage rate r becomes P + r P = P ( 1 + r)
Example 1: 200 increased by 5% becomes 200 + (5/100)*200 = 200(1 + 0.05)
Simple Interest (not compounded)
If P is deposited in a saving account at an interest r, that is not compounded, for t years then the interest I earned after t years is given by
I = P r t
and the total amount A in the account after t years is given by
A = P + P r t = P(1 + r t)
Example 1: $100 in a 3% interest rate saving account, not compounded, would earns 100*(3/100)*5 = $15 in 5 years.
Yearly Interest Compounding (Savings Account for Example)
An amount of money P (principal) is invested at an annual percentage rate r. What is the total amount of money A after t years?
 At t = 0 , A = P (the day money is deposited)

At the end of the first year t = 1 , A = P + r P = P(1 + r)
we have used what was explained above: A quantity P increased by a percentage r becomes P + r P = P ( 1 + r)

At the end of the second year t = 2 , A = P(1 + r) + r P(1 + r)
we have used what was explained above: A quantity P ( 1 + r) (at the end of year 1) increased by by a percentage r becomes P ( 1 + r) + r P ( 1 + r)
which gives
A = (P + r P)(1 + r) = P(1 + r )(1 + r) = P(1 + r)^{ 2}

At the end of the third year t = 3 , A = P(1 + r)^{ 2} + r P(1 + r)^{ 2} = P(1 + r)^{ 2}(1 + r) = P(1 + r)^{ 3}

At the end of the fourth year t = 4 , A = P(1 + r)^{ 3} + rP(1 + r)^{ 3} = P(1 + r)^{ 3}(1 + r) = P(1 + r)^{ 4}

At the end of year t, A = P(1 + r)^{ t} by extension of the above
So if an amount P (principal) is invested at the annual rate r and is compounded annually, the total amount A at the end of t years is given by
A = P(1 + r)^{ t}
Example 2: $1000 is invested for 3 years, compounded every year, at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)
A = P(1 + r)^{ t} = 1000(1 + 0.03)^{ 3} = $1092.73
You may use the calculator to input and experiment with more values for P, r and t and obtain the amount A. Use your own calculator and compare the results.
Interest Compounding n Times Per Year
How about compounding more that once a year? Let us say the interest is compounded twice a year (every 6 months) as follows:
Yearly rate is r; set a half yearly rate equal to r/2 and compound twice a year as follows:
t = 0 , A = P
At the end of the first 6 months of the year: A = P(1 + r/2)
At the end of the second 6 months of the same year: A = P(1 + r/2) + (r/2) P(1 + r/2) = P(1 + r/2)^{ 2}
We now extend the above to write:
So if an amount P (principal) is invested at the annual rate r and is compounded n times a year , the amount at the end of t years is given by
A = P(1 + r/n)^{ n t}
Example 3: $1000 is invested for 3 years, compounded twice a year (n = 2), at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)
A = P(1 + r/n)^{ n t} = 1000(1 + 0.03 / 2)^{ 2×3} = $1093.44
You may use the calculator to input and experiment with more values for P, r, t and n and obtain the amount A. Use your own calculator and compare the results.
To understand the advantage of Compounding more than once a year, Keep P, r and t constant (The same amount invested at the rate r for t years) and increase n. What happens to A?
Continuous Compounding
So if an amount P (principal) is invested at the annual rate r and is compounded n times a year , the amount at the end of t years is given by (see above)
A = P(1 + r/n)^{ n t}
Let N = n / r , then r / n = 1 / N and n = r N , hence the formula for A becomes
A = P(1 + 1 / N)^{ N r t}
Which can be written as
A = P ( (1 + 1 / N)^{ N } ) ^{ r t}
The question that one may ask is that what if we increase n indefinitely, which means increasing N indefinitely in our formula?
Use the calculator below to increase N ( 2 ,12, 1000, 10000...) and see how the value of the term (1 + 1 / N)^{ N } behaves?
As the number of compounding n increases, N also increases, the term (1 + 1 / N)^{ N } approaches a constant value which is called e and is approximately equal 2.718282... . More rigously, e is defined as the limit of (1 + 1/N)^{ N} as N approaches infinity.
Hence for continuous compounding (n very large) at the rate r and an initial amount P and after t years A is given by:
A = P e ^{ r t}
Example 4: $1000 is invested for 3 years, compounded continuously, at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)
A = P e ^{ r t} = 1000 e ^{ 0.03 × 3} = $1094.17
You may use the calculator to input and experiment with more values for P, r, and obtain the amount A. Use your own calculator and compare the results.
Conclusion: compare the way the same amount of $1000 was compounded in the eaxamples 1,2 and 4 and make a conclusion as to which compounding earns more.