4 12 Continuous Compounding 1 365

Given: P=$2500, r=6%, t=20 years

The formula for Compound Interest is:

$A=P{\left(1+\frac{{\displaystyle \frac{r}{n}}}{100}\right)}^{tn}$, where A is the amount, P is the principal, r is the rate of interest, n is the number of times compounding is done in a year and t is the time period.

For n=1:

$\begin{array}{rcl}A& =& P{\left(1+\frac{{\displaystyle \frac{r}{n}}}{100}\right)}^{tn}\\ & =& 2500{\left(1+\frac{{\displaystyle 6}}{100}\right)}^{20}\\ & =& 2500{\left(1.06\right)}^{20}\\ & =& 8,017.83\end{array}$

Thus, when n=1 the value of the amount is $8,017.83

For n=2:

$\begin{array}{rcl}A& =& P{\left(1+\frac{{\displaystyle \frac{r}{n}}}{100}\right)}^{tn}\\ & =& 2500{\left(1+\frac{{\displaystyle \frac{{\displaystyle 6}}{2}}}{100}\right)}^{20\times 2}\\ & =& 2500{\left(1+\frac{{\displaystyle 3}}{100}\right)}^{40}\\ & =& 2500{\left(1.03\right)}^{40}\\ & =& 8,155.09\end{array}$

Thus, when n=2 the value of the amount is $8,155.09.

For n=4:

$\begin{array}{rcl}A& =& P{\left(1+\frac{{\displaystyle \frac{r}{n}}}{100}\right)}^{tn}\\ & =& 2500{\left(1+\frac{{\displaystyle \frac{{\displaystyle 6}}{4}}}{100}\right)}^{20\times 4}\\ & =& 2500{\left(1+\frac{{\displaystyle 1.5}}{100}\right)}^{80}\\ & =& 2500{\left(1.015\right)}^{80}\\ & =& 8,226.65\end{array}$

Thus, when n=4 the value of the amount is $8,226.65.