Prove the following by using the principle of mathematical induction for all $n \in N$:
$a + ar + ar^{2} + \ldots + ar^{n-1} = \frac{a(r^{n} - 1)}{r - 1}$

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): a + ar + ar^{2} + \ldots + ar^{n-1} = \frac{a(r^{n} - 1)}{r - 1}$
For $n = 1$,we have
$P(1): a = \frac{a(r^{1} - 1)}{r - 1} = a$,which is true.
Let $P(k)$ be true for some positive integer $k$,i.e.,
$a + ar + ar^{2} + \ldots + ar^{k-1} = \frac{a(r^{k} - 1)}{r - 1}$ $(i)$
We shall now prove that $P(k+1)$ is true.
Consider the sum of the first $(k+1)$ terms:
$(a + ar + ar^{2} + \ldots + ar^{k-1}) + ar^{(k+1)-1}$
$= \frac{a(r^{k} - 1)}{r - 1} + ar^{k}$ [Using $(i)$]
$= \frac{a(r^{k} - 1) + ar^{k}(r - 1)}{r - 1}$
$= \frac{ar^{k} - a + ar^{k+1} - ar^{k}}{r - 1}$
$= \frac{ar^{k+1} - a}{r - 1}$
$= \frac{a(r^{k+1} - 1)}{r - 1}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.