A. The amount (A) at the end of t years of continuous compounding of principal P at rate r will be
... A = Pe^(rt)
For P=1000, r=.02, and t=1, The amount is
... A = $1000e^(.02·1) = $1020.20134
B. The formula for daily compounding is
... A = P(1 + r/365)^(365t)
Using the same values of P, r, t, the amount is
... A = $1000(1 +.02/365)^365 = $1020.20078
Continuous compounding produces a larger result.
The result gets larger the more often compounding occurs. Continuous compounding is the highest possible rate at which compounding can take place, so produces the largest possible result.
C. The balance at the end of the year when interest is compounded n times per year is given by
... A = P(1 + r/n)^n
Each year interest is compounded this way, the amount is multiplied by
... (1 + r/n)^n
When this happens each year for t years, the multiplier has been applied t times. Exponentiation is used to represent the effect of such repeated multiplication, so the balance at the end of t years is
... A = P((1 + r/n)^n)^t = P(1 +r/n)^(nt)
D. (Note the previous answer assumed the existence of this answer.) The same logic as for C above applies for each period that compounding takes place. That is, if compounding occurs n times per year, the interest rate applied for each period is the nominal annual rate r divided by the number of periods n. The multiplier applied to the initial principal amount is
... (1 + r/n)
When than factor is used n times during the year, the multiplier of the initial principal amount is
... (1 + r/n)·(1 + r/n)· ... ·(1 + r/n) . . . where the factor is applied n times.
In more compact notation, this multiplier is
... (1 +r/n)^n
When that multiplier is applied to principal P, the account balance A at the end of the year is ...
... A = P(1 +r/n)^n
