Let the given statement be P(n), i.e.,
1 . 3 + 2 . 32 + 3 . 33 + ... + n.3n =

(2n−1)3n+1+3

4

First we prove that the statement is true for n = 1.
P (n) : 1 . 3 =

(2.1−1)31+1+3

4

⇒ 3 =

32+3

4

=

9+3

4

=

12

4

= 3
which is true.
Assume P(k) is true for some positive integer k, i.e.,
1 . 3 + 2 . 32 + 3 . 33 + ... + k.3k =

(2k−1)3k+1+3

4

... (i)
We shall now prove that P(k + 1) is also true.
For this we have to prove that
1 . 3 + 2 . 32 + 3 . 33 + ... + k.3k+(k+1).3k+1 =

[2(k+1)−1]3k+2+3

4

L.H.S. = 1 . 3 + 2 . 32 + 3 . 33 + ... + k.3k+(k+1).3k+1
=

(2k−1)3k+1+3+4(k+1)3k+1

4

[From (i)]
=

(2k−1+4k+4)3k+1+3

4

=

(6k+3)3k+1+3

4

=

[2(k+1)−1]3k+1+1+3

4

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