Let the given statement be P(n), i.e.,
P (n) : a + ar + ar2 + ... + arn−1 =

a(rn−1)

r−1

First we prove that statement is true for n = 1.
P (1) : a =

a(r1−1)

r−1

= a , which is true
Assume P(k) is true for some positive integer k, i.e.,
a + ar + ar2 + ... + ark−1 =

a(rk−1)

r−1

... (i)
Now prove that P(k + 1) is also true.
For this we have to prove that
a + ar + ar2 + ... + ark−1+ar(k+1)−1 =

a(rk+1−1)

r−1

L.H.S. = a + ar + ar2 + ... + ark−1+ar(k+1)−1
=

a(rk−1)

r−1

+ar(k+1)−1 From (i)
=

ark−a+ark(r−1)

r−1

= ark−a+ark+1−a

rk

r−1

=

ark+1−a

r−1

=

a(kk+1−1)

r−1

= R.H.S.
Thus, P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.