a + ar + ar


$a+a r+a r^{2}+\ldots+a r^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right), r \neq 1$


Let P(n) be the given statement.


Step 1;

$P(n)=a+a r+a r^{2}+\ldots+a r^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right), r \neq 1$


Hence, $P(1)$ is true.

Step 2 :


Suppose $P(m)$ is true.


$a+a r+a r^{2}+\ldots+a r^{m-1}=a\left(\frac{r^{m}-1}{r-1}\right), r \neq 1$

To show: $P(m+1)$ is true whenever $P(m)$ is true.

That is,

$a+a r+a r^{2}+\ldots+a r^{m}=a\left(\frac{r^{m+1}-1}{r-1}\right), r \neq 1$

$W e$ know that $P(m)$ is true.


Thus, we have :

$a+a r+a r^{2}+\ldots+a r^{m-1}=a\left(\frac{r^{m}-1}{r-1}\right)$

$\Rightarrow a+a r+a r^{2}+\ldots+a r^{m-1}+a r^{m}=a\left(\frac{r^{m}-1}{r-1}\right)+a r^{m} \quad$ [Adding $a r^{m}$ to both sides]

$\Rightarrow P(m+1)=a\left(\frac{r^{m}-1+r \cdot r^{m}-r^{m}}{r-1}\right)$


$\Rightarrow P(m+1)=a\left(\frac{r^{m+1}-1}{r-1}\right), r \neq 1$

Thus, $P(m+1)$ is true.


By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.

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