We now apply our knowledge of the simple interest model to build the compound interest model.
Half-yearly compound interest
Let us first consider the scenario of interest compounded half-yearly.
A year comprises of $2$ half-years (i.e., $2$ periods of six months). Assume that the bank pays out a simple interest over each of these half-years at a rate of $R$% per annum. Then on each half-year, the (simple) interest would be pay out at $\frac{R}{2}$% per half-yearly.
At the end of the first year, the amount accumulated would be $P \left(1+ \frac{R}{100}\right)$ dollars.
This amount is viewed as the principal sum at the start of the second half-year. So, the amount generated over the second half-year would be $P \left(1+ \frac{R}{100}\right)^2$ dollars.
Half-yearly compound interest
Principal = $\$ P$ Interest rate = $R \%$ per annum Compounded half-yearly Time = $n$ years Compounded amount = $\$A$ Then 
\[A=P \left(1+ \frac{R}{2\times 100}\right)^{2n}\]
Total compound interest $I$ compounded half-yearly over $n$ years is given by
\[I=A-P=P\left[\left(1+ \frac{R}{2(100)}\right)^{2n} -1\right]\]
Quarterly compound interest
A year comprises of $4$ quarters (i.e., $4$ periods of three months). Assume that the bank pays out a simple interest over each of these quarters at a rate of $R$% per annum. Then on each quarter, the (simple) interest would be pay out at $\frac{R}{4}$% per annum. At the end of the first year, the amount accumulated would be $P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)$ dollars. This amount is viewed as the principal sum at the start of the second half-year. So, the amount generated over the second quarter would be $P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)^2$ dollars.
Repeating the argument twice, the amount generated over the fourth quarter is $P \left( 1 + \frac{\frac{R}{4}}{100} \right)^4$.
Quarterly compound interest
Principal = $$P$ Interest rate = $R$% per annum Compounded quarterly Time = $n$ years Compounded amount = $$A$ Then
\[A = P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)^{4n}\]
Total compound interest $I$ compounded quarterly over $n$ years is given by
\[I = A - P = P \left[ \left( 1 + \frac{R}{4(100)} \right)^{4n} - 1 \right]\]
In general, if a bank pays out interest compounded per $\frac{1}{k}$ of a year over $n$ years, then the compound interest paid out can be shown to be
\[I = P \left[ \left( 1 + \frac{R}{100k} \right)^{kn} - 1 \right]\]
Remarks. Note that over each frame of $\frac{1}{k}$-year, the amount accumulated has dilated by a factor of $\left( 1 + \frac{R}{100k} \right)$, which can be viewed as a common ratio. This common ratio is multiplied to the current amount to yield the next amount. In this way, what lies in the heart of a compound interest model is the geometric progression.

We now apply our knowledge of the simple interest model to build the compound interest model.

Let us first consider the scenario of interest compounded half-yearly.

A year comprises of $2$ half-years (i.e., $2$ periods of six months). Assume that the bank pays out a simple interest over each of these half-years at a rate of $R$% per annum. Then on each half-year, the (simple) interest would be pay out at $\frac{R}{2}$% per half-yearly.

At the end of the first year, the amount accumulated would be $P \left(1+ \frac{R}{100}\right)$ dollars.

This amount is viewed as the principal sum at the start of the second half-year. So, the amount generated over the second half-year would be $P \left(1+ \frac{R}{100}\right)^2$ dollars.

Principal = $\$ P$ 
Interest rate = $R \%$ per annum
Compounded half-yearly
Time = $n$ years
Compounded amount = $\$A$
Then 

\[A=P \left(1+ \frac{R}{2\times 100}\right)^{2n}\]

Total compound interest $I$ compounded half-yearly over $n$ years is given by

\[I=A-P=P\left[\left(1+ \frac{R}{2(100)}\right)^{2n} -1\right]\]

A year comprises of $4$ quarters (i.e., $4$ periods of three months). Assume that the bank pays out a simple interest over each of these quarters at a rate of $R$% per annum. Then on each quarter, the (simple) interest would be pay out at $\frac{R}{4}$% per annum. At the end of the first year, the amount accumulated would be $P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)$ dollars. This amount is viewed as the principal sum at the start of the second half-year. So, the amount generated over the second quarter would be $P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)^2$ dollars.

Repeating the argument twice, the amount generated over the fourth quarter is $P \left( 1 + \frac{\frac{R}{4}}{100} \right)^4$.

Principal = $$P$
Interest rate = $R$% per annum
Compounded quarterly
Time = $n$ years
Compounded amount = $$A$
Then

\[A = P \left( 1 + \frac{\big(\frac{R}{4}\big)}{100} \right)^{4n}\]

Total compound interest $I$ compounded quarterly over $n$ years is given by

\[I = A - P = P \left[ \left( 1 + \frac{R}{4(100)} \right)^{4n} - 1 \right]\]

In general, if a bank pays out interest compounded per $\frac{1}{k}$ of a year over $n$ years, then the compound interest paid out can be shown to be

\[I = P \left[ \left( 1 + \frac{R}{100k} \right)^{kn} - 1 \right]\]

**Remarks.** Note that over each frame of $\frac{1}{k}$-year, the amount accumulated has dilated by a factor of $\left( 1 + \frac{R}{100k} \right)$, which can be viewed as a common ratio. This common ratio is multiplied to the current amount to yield the next amount. In this way, what lies in the heart of a compound interest model is the geometric progression.

Compound interest

Compound interest