Question

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

Answer 1

Let P(n) : $a+ar+a{r}^2+....+a{r}^{n-1}=a\left(\frac{r^n-1}{r-1}\right)$, r $\ne$ 1.

For n = 1

a = a $\left(\frac{r^1-1}{r-1}\right)$

a = a

$\Rightarrow$ p(n) is true for n = 1

Let p(n) is true for n = k, so

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

We have to show that,

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

Now,

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

[using equation (i)]

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

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

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

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ is true for all n $ϵ$ N by PMI