Prove by induction $a+ar++a{r}^2+.....$ upto $n$ terms $=\frac{a\left(r^n-1\right)}{r-1},r\ne 1$

$n^{th}$ term is $a{r}^{n-1}$

$S\left(n\right)=a+ar+a{r}^2+\dots \dots \dots +a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$

Let $S\left(n\right)$ be the given statement

Put $n=1$

L.H.S $=a{r}^{1-1}=a{r}^0=a$

R.H.S $=\frac{a(r^1-1)}{r-1}=a$

L.H.S = R.H.S

$S\left(1\right)$is true

Assume that $S\left(k\right)$is true for some $k\in n$

$a+ar+a{r}^2+\dots \dots +a{r}^{k-1}=\frac{a(r^k-1)}{r-1}$

Adding ${(k + 1)}^{th}$ terms on both sides

$a+ar+a{r}^2+\dots .+a{r}^{k-1}+a{r}^k=\frac{a\left(r^k-1\right)}{r-1}+a{r}^k$

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

$=\frac{a{r}^k-a+a{r}^{k}r-a{r}^k}{r-1}=\frac{a{r}^{k+1}-a}{r-1}$

$=\frac{a\left(r^{k+1}-1\right)}{r-1}$    $\therefore S(k+1)$ is true

By the principle of finite mathematical induction
$S\left(n\right)$ is true for all $n\in N$