Let the given statement be P(n), i.e.,
P (n) : 1 +

1

(1+2)

+

1

(1+2+3)

+ ... +

1

(1+2+3+...+n)

=

2n

(n+1)

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

2×1

(1+1)

=

2

2

= 1 , which is true
Assume that P(k) is true for some positive integer k, i.e.,
1 +

1

(1+2)

+

1

(1+2+3)

+ ... +

1

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

=

2k

(k+1)

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

1

(1+2)

+

1

(1+2+3)

+ ... +

1

(1+2+...+(k+1))

=

2(k+1)

((k+1)+1)

L.H.S. = 1 +

1

(1+2)

+

1

(1+2+3)

+ ... +

1

(1+2+...+(k+1))

= 1 +

1

(1+2)

+

1

(1+2+3)

+ ... +

1

(1+2+...+k)

+

1

(1+2+...+(k+1))

=

2k

(k+1)

+

1

(k+1)(k+2) 2

[From (i)]
=

2k

(k+1)

+

2

(k+1)(k+2)

=

2k(k+2)+2

(k+1)(k+2)

=

2(k2+2k+1)

(k+1)(k+2)

=

2(k+1)

(k+2)

= 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.