A)  Interest credited once a year. (n=1).

If P  dollars are deposited at r%, then the value at the end of one year is

P+rP or P(1+r), where r% is the annual rate expressed as a decimal.


B)  Interest credited twice a year. (n=2).

The interest rate every 6 months is (r/2)%.
The value at the end of one year is

P+(r/2)P+(r/2)[P+(r/2)P]=[P+(r/2)P][1+(r/2)]=P[1+(r/2)][1+(r/2)]=
P[1+(r/2)]²


C)  Interest credited 4 times a year. (n=4).

The interest rate every 3 months is (r/4)%.

The value at the end of one year is P+(r/4)P+(r/4){[P+(r/4)P]}+(r/4){[P+(r/4)P]+(r/4)[(P+(r/4)P]}+(r/4){[P+(r/4)P]+(r/4)[(P+(r/4)P]+(r/4)[(P+(r/4)P+(r/4)[P+(r/4)P]}=[P+(r/4)P]{1+(r/4)+(r/4)[1+(r/4)]+(r/4)[1+(r/4)+(r/4)(1+(r/4))]}=P[(1+(r/4)(1+(r/4)[1+(r/4)+(r/4)+(r/4)]=P[1+(r/4)][1+(r/4]{[1+(r/4)]+(r/4)[1+(r/4)]}
= P[1+(r/4)][1+(r/4)][1+(r/4)][1+(r/4)]=P[1+(r/4)]⁴

D)  Generally, interest credited n times a year would yield a total value at the end of one year of

P[1+(r/n)]ⁿ

AFTER t YEARS

If P earns interest, compounded n times/year, the value will be  P[1+(r/n)]ⁿ at the end of one year.

At the end of 2 years, the value will be  P[1+(r/n)]ⁿ[1+(r/n)]ⁿ= P[1+(r/n)]⁽²ⁿ⁾

At the end of 3 years, the value will be  P[1+(r/n)]ⁿ[1+(r/n)]ⁿ[1+(r/n)]ⁿ=
P[1+(r/n)]⁽³ⁿ⁾

Continuing, we can see, the value at the end of t years would be  
P[1+(r/n)]^(tn)

Continuous Compounding

We've seen how interest is credited compounded a number of times over t years. If the interest is compounded at every instant of time, we have continuous compounding. Using the calculus limit definition of the number e (base for natural logs), we come up with the formula for continuous growth or, in the case of interest, continuous compound interest.  

Let P_o= the original amount of principle (amount deposited).
Let the interest rate be r% (expressed as a decimal).
Then the principle will grow, at this continuous compounding rate according to the formula:

P = P_oe^(rt)