Let the given statement be P(n), i.e.,
P (n) :

1

1.4

+

1

4.7

+

1

7.10

+ ... +

1

(3n−2)(3n+1)

=

n

(3n+1)

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

1

1.4

=

1

3.1+1

⇒

1

4

=

1

4

, which is true
Assume P(k) is true for some positive integer k, i.e.,

1

1.4

+

1

4.7

+

1

7.10

+ ... +

1

(3k−2)(3k+1)

=

k

(3k+1)

... (i)
Now we shall prove that P(k + 1) is also true.
For this we have to prove that

1

1.4

+

1

4.7

+

1

7.10

+ ... +

1

(3k−2)(3k+1)

+

1

(3(k+1)−2)(3(k+1)+1)

=

(k+1)

3(k+1)+1

L.H.S. =

1

1.4

+

1

4.7

+

1

7.10

+ ... +

1

(3k−2)(3k+1)

+

1

(3(k+1)−2)(3(k+1)+1)

=

k

3k+1

+

1

(3k+1)(3k+4)

[From (i)]
=

1

3k+1

[k+

1

3k+4

] =

1

3k+1

[

3k2+4k+1

3k+4

]
=

(3k+1)(k+1)

(3k+1)(3k+4)

=

(k+1)

3k+4

=

(k+1)

3(k+1)+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.