Let P(n): 2 + 5 + 8 + 11 + … + (3n – 1) = n(3n + 1)

For n=1

P(1): 2 = .1.(4)

2 = 2

Since, P(n) is true for n = 1

Let P(n) is true for n = k, so

P(k): 2 + 5 + 8 + 11 + … + (3k – 1) = k(3k + 1) - - - - - - - (1)

We have to show that,

2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2) = (k + 1)(3k + 4)

Now,

{2 + 5 + 8 + 11 + … + (3k – 1)} + (3k + 2)

= .k(3k + 1) + (3k + 2)

=

=

=

=

=

=

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

Hence, P(n) is true for all n ∈ N by PMI