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

1

2.5

+

1

5.8

+

1

8.11

+ ... +

1

(3n−1)(3n+2)

=

n

(6n+4)

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

1

2.5

=

1

(6.1+4)

⇒

1

10

=

1

10

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

1

2.5

+

1

5.8

+

1

8.11

+ ... +

1

(3k−1)(3k+2)

=

k

(6k+4)

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

1

2.5

+

1

5.8

+

1

8.11

+ ... +

1

(3k−1)(3k+2)

+

1

(3k+2)(3k+5)

=

k+1

6(k+1)+4

L.H.S. =

1

2.5

+

1

5.8

+

1

8.11

+ ... +

1

(3k−1)(3k+2)

+

1

(3k+2)(3k+5)

=

k

6k+4

+

1

(3k+2)(3k+5)

[From (i)]
=

1

(3k+2)

[

k

2

+

1

3k+5

] =

1

(3k+2)

[

3k2+5k+2

2(3k+5)

]
=

3k2+5k+2

2(3k+2)(3k+5)

=

(3k+2)(k+1)

2(3k+2)(3k+5)

=

k+1

6k+10

=

k+1

6(k+1)+4

= R.H.S.
Since k ≠ -2/3
Thus P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true ∀ n ∈ N.