Chapter 10.1, Problem 52E

a.

To determine

To find: Value of $S-rS$

Answer to Problem 52E

  $S-rS=a-a{r}^n$

Explanation of Solution

Given:

  $S=a+ar+a{r}^2+\mathrm{...}+a{r}^{n-1}$

Here, a and r are real numbers and $r\ne 1$

Calculation:

As, $S=a+ar+a{r}^2+\mathrm{...}+a{r}^{n-1}$

Multiply r on both side

  $rS=ar+a{r}^2+a{r}^3+\mathrm{.....}+a{r}^n$

So,

  $\begin{array}{l}S-rS=S=a+ar+a{r}^2+\mathrm{...}+a{r}^{n-1}-\left(rS=ar+a{r}^2+a{r}^3+\mathrm{.....}+a{r}^n\right)\\ \Rightarrow S-rS=a-a{r}^n\end{array}$

b.

To determine

To show: $S=\frac{a-a{r}^n}{1-r}$

Explanation of Solution

Given:

  $S=a+ar+a{r}^2+\mathrm{...}+a{r}^{n-1}$

Calculation:

As from the previous subpart,

  $S-rS=a-a{r}^n$

So,

  $\begin{array}{l}S\left(1-r\right)=a-a{r}^n\\ \Rightarrow S=\frac{a-a{r}^n}{1-r}\end{array}$

Hence it is proved that $S=\frac{a-a{r}^n}{1-r}$