a + ar + ar2 + … + arn – 1 = a [(rn – 1)/(r – 1)], r ≠ 1

Let us check for n = 1,

P (1): a =\(\dfrac{ a (r^1 – 1)}{(r-1)}\)

a = a

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): a + ar + ar² + … + ar^{k – 1} = \(\dfrac{ a (r^k – 1)}{(r – 1)}\) … (i)

So,

a + ar + ar² + … + ar^{k – 1} + ar^k

Now, substituting the value of P (k) we get,

= \(\dfrac{a(rk – 1)}{(r – 1)}\) + ar^k by using equation (i)

=\( \dfrac{ a(rk+1 – 1)}{ (r-1)}\)

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.