Question

Show that: $a+ar+a{r}^2+...+a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$

Answer 1

Let sum of the series $a+ar+a{r}^2+........+a{r}^{n-1}$ is $S$

$S=a+ar+a{r}^2+........+a{r}^{n-1}$ ..................... (1)

$rS=ar+a{r}^2+........+a{r}^{n-1}+a{r}^n$ .................. (2)

Subtract (1) from (2),

$rS-S=a{r}^n-a$

$S(r-1)=a(r^n-1)$

$S=\frac{a(r^n-1)}{r-1}$