Chapter 8.3, Problem 97E

To determine

To find: the value of $A$ .

Answer to Problem 97E

The value of $A$ is $6480.83.

Explanation of Solution

Given information:

A deposit of $100 is made at the beginning of given month in an account that pays 3% interest, compounded monthly. The balance $A$ in the account at the end of 5 years is given by

  $A=100{\left(1+\frac{0.03}{12}\right)}^1+\mathrm{...}+100{\left(1+\frac{0.03}{12}\right)}^{60}$

Formula used:

The sum $S_n$ or ${\displaystyle \sum _{i=1}^n{a}_r{r}^{r-1}}$ of a finite geometric sequence is $a_1\left(\frac{1-r^n}{1-r}\right)$ , where $a_1$ is the first term, $r$ is the common ration, and $n$ is the number of terms in the sequence.

Calculation:

Now determine the net balance $A$ in the account as shown below:

  $\begin{array}{c}A=100{\left(1+\frac{0.03}{12}\right)}^1+\mathrm{...}+100{\left(1+\frac{0.03}{12}\right)}^{60}\\ =100{\left(\frac{12.03}{12}\right)}^1+\mathrm{...}+100{\left(\frac{12.03}{12}\right)}^{60}\\ =100\left(\frac{12.03}{12}\right)\left\{1+{\left(\frac{12.03}{12}\right)}^1+{\left(\frac{12.03}{12}\right)}^2+\mathrm{...}+{\left(\frac{12.03}{12}\right)}^{59}\right\}\\ =100.25\left\{1+{\left(1.0025\right)}^1+{\left(1.0025\right)}^2+\mathrm{...}+{\left(1.0025\right)}^{59}\right\}\end{array}$

Continue simplification

  $\begin{array}{c}\left\{{\displaystyle \sum _{n=1}^{60}100.25{\left(1.0025\right)}^{n-1}}\right\}\text{Use}\text{sigma}\text{notation}\\ =100.25\left(\frac{{\left(1.0025\right)}^{60}-1}{\left(1.0025\right)-1}\right)\\ =6480.8329403677\\ =6480.83\\ A=\$6480.83\end{array}$