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

1

1.2.3

+

1

2.3.4

+

1

3.4.5

+ ... +

1

(n+1)(n+2)

=

n(n+3)

4(n+1)(n+2)

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

1

1.2.3

=

1.(1+3)

4(1+1)(1+2)

⇒

1

6

=

4

4.2.3

=

1

6

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

1

1.2.3

+

1

2.3.4

+

1

3.4.5

+ ... +

1

(k+1)(k+2)

=

k(k+3)

4(k+1)(k+2)

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

1

1.2.3

+

1

2.3.4

+

1

3.4.5

+ ... +

1

(k+1)(k+2)

+

1

(k+1)(k+2)(k+3)

=

(k+1)9k+4)

4(k+2)(k+3)

L.H.S. =

1

1.2.3

+

1

2.3.4

+

1

3.4.5

+ ... +

1

(k+1)(k+2)

+

1

(k+1)(k+2)(k+3)

=

k(k+3)

4(k+1)(k+2)

+

1

(k+1)(k+2)(k+3)

[From (i)]
=

1

(k+1)(k+2)

[

k(k+3)

4

+

1

(k+3)

] =

1

(k+1)(k+2)

[

k(k+3)2+4

4(k+3)

]
=

k(k2+6k+9)+4

4(k+1)(k+2)(k+3)

=

k3+6k2+9k+4

4(k+1)(k+2)(k+3)

=

(k+1)(k2+5k+4)

4(k+1)(k+2)(k+3)

=

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

4(k+1)(k+2)(k+3)

=

(k+1)(k+2)

4(k+2)(k+3)

Thus, P(k + 1) is true, whenever P(k) is true.
Hence, by the principal of mathematical induction P(n) is true.