Chapter 5.4, Problem 64ES

Let

$S={\displaystyle \underset{k=0}{\overset{n}{\sum }}a{r}^k}$

Show that $S-rS=a-a{r}^{n+1}$ and hence that

${\displaystyle \underset{k=0}{\overset{n}{\sum }}a{r}^k}=\frac{a-a{r}^{n+1}}{1-r}\left(r\ne 1\right)$

(A sum to this form is called a geometric sum.)