Question

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

Answer 1

Let P(n) $a+ar+a{r}^2+\cdots +a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$
For n = 1
$P\left(1\right)=a{r}^{1-1}=\frac{a(r^1-1)}{r-1}\Rightarrow a=a\therefore$ P(1) is true
Let P (n) be true for n = k
$\therefore P\left(k\right)=a+ar+a{r}^2+\cdots +a{r}^{k-1}=\frac{a(r^k-1)}{r-1}$
For n = k + 1
$R.H.S.=\frac{a(r^{k+1}-1)}{r-1}L.H.S.=\frac{a(r^k-1)}{r-1}+a{r}^k=\frac{a{r}^k}{r-1}-\frac{a}{r-1}+a{r}^k=a{r}^k(\frac{1}{r-1}+1)-\frac{a}{r-1}=a{r}^k.\left(\frac{r}{r-1}\right)-\frac{a}{r-1}=\frac{a{r}^{k+1}}{r-1}-\frac{a}{r-1}=\frac{a{r}^{k+1}-a}{r-1}=\frac{a(r^{k+1}-1)}{r-1}$
$\therefore$ P(k +1) is true
Thus P (k) is true $\Rightarrow P(k+1)$ is true
Hence by principle fo mathematical induction,
P(n) is true for all $nϵN.$