Chapter 9.3, Problem 85E

To determine

Show that balance is equal to the given.

Answer to Problem 85E

The sum of a finite geometric sequence is $P\left[{\left(1+\frac{r}{12}\right)}^{12r}-1\right]\left(1+\frac{12}{r}\right)$

Explanation of Solution

Given:

The balance A in account at the end of t years is,

  $A=P{\left(1+\frac{r}{12}\right)}^1+P{\left(1+\frac{r}{12}\right)}^2+\cdots +P{\left(1+\frac{r}{12}\right)}^{12r}$

The balance A in account at the end of t years is,

  $A=P{\left(1+\frac{r}{12}\right)}^1+P{\left(1+\frac{r}{12}\right)}^2+\cdots +P{\left(1+\frac{r}{12}\right)}^{12r}$

  $\begin{array}{l}{a}_1=P\left(1+\frac{r}{12}\right)\hfill \\ r=\left(1+\frac{r}{12}\right)\hfill \\ n=12t\hfill \end{array}$

Now, let's apply the formula for the sum of a finite geometric sequence,

  $\begin{array}{cc}A& =S_m\\ & =a_1\left(\frac{1-r^{\ast }}{1-r}\right)\\ & =P\left(1+\frac{r}{12}\right)\left(\frac{1+\frac{r^{12}}{12}}{1+\frac{r}{12}}\right)\\ & =P\left[{\left(1+\frac{r}{12}\right)}^{12r}-1\right]\left(1+\frac{12}{r}\right)\end{array}$

The sum of a finite geometric sequence is $P\left[{\left(1+\frac{r}{12}\right)}^{12r}-1\right]\left(1+\frac{12}{r}\right)$