Chapter 8.3, Problem 100E

To determine

To show: the balance is given by $A=\frac{P{e}^{r/12}\left(e^{rt}-1\right)}{e^{r/12}-1}$ .

Explanation of Solution

Given information:

A deposit of P dollars is made at the beginning of given month in an account earning an annual interest rate $r$ , compounded continuously. The balance A after $t$ years is given by

  $A=P{e}^{r/12}+P{e}^{2r/12}+\mathrm{...}+P{e}^{12tr/12}$

Calculation:

The balance $A$ in the account at the end of 5 years can be calculated by using the formula sum of finite geometric progression.

In general, the sum of the finite geometric sequence $a_1,a_{1}r,a_1{r}^2,a_1{r}^3,a_1{r}^4,\mathrm{...},a_1{r}^{n-1}$ with common ratio $r\ne 1$ is as follows,

  $S_n={\displaystyle \sum _{i=1}^n{a}_1{r}^{i-1}=a_1\left(\frac{1-r^n}{1-r}\right),r}\ne 1$

Substitute the values $a_1=P{e}^{\frac{r}{12}},r=e^{\frac{r}{12}},n=12t$ in the above formula and obtain the balance $A$ in the account monthly as,

  $\begin{array}{c}e{S}_t=P{e}^{\frac{r}{12}}\left(\frac{1-{\left(e^{\frac{r}{12}}\right)}^{12t}}{1-\left(e^{\frac{r}{12}}\right)}\right)\\ =P{e}^{\frac{r}{12}}\left(\frac{1-e^n}{1-e^{\frac{r}{12}}}\right)\\ =\frac{P{e}^{\frac{r}{12}}\left(e^{rt}-1\right)}{\left(e^{\frac{r}{12}}-1\right)}\end{array}$

Hence, it is proved that the balance $A$ in the account monthly is

  $\frac{P{e}^{\frac{r}{12}}\left(e^{rt}-1\right)}{\left(e^{\frac{r}{12}}-1\right)}$ .