Let the given statement be P(n) i.e.,
P (n) : 1 + 3 + 32 + ... + 3n−1 =

(3n−1)

2

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

31−1

2

=

2

2

= 1 , which is true
Assume P(k) is true for some positive integers k, i.e.,
1 + 3 + 32 + ... + 3k−1 =

3k−1

2

... (i)
We shall now prove that P(k + 1) is also true.
For this we have to prove that
1 + 3 + 32 + ... + 3k−1+3k =

(3k+1−1)

2

By adding 3k to both the sides of (i), we get
L.H.S. = 1 + 3 + 32 + ... + 3k−1+3k =

3k−1

2

+3k [from (i)]
=

3k−1+2.3k

2

=

3.3k−1

2

=

3k+1−1

2

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