Let the given statement be P(n), i.e.,
P (n) : 1 + 2 + 3 + ... + n <

1

8

(2n+1)2
First we prove that the statement is true for n = 1.
P (1) : 1 <

1

8

(2.1+1)2 =

1

8

(3)2 =

9

8

⇒ 1 <

9

8

, which is true
Assume P(k) is true for some positive integer k, i.e.,
1 + 2 + 3 + ... + k <

1

8

(2k+1)2 ... (i)
Now we shall prove that P(k + 1) is true.
For this we have to prove that
1 + 2 + 3 + ... + k + (k + 1) <

1

8

[2(k+1)+1]2
L.H.S. = 1 + 2 + 3 + ... + k + (k + 1) <

1

8

(2k+1)2 + (k + 1) [From (i)]
=

4k2+4k+1+8k+8

8

=

4k2+12k+9

8

=

(2k+3)2

8

=

[2(k+1)+1]2

8

= R.H.S.
⇒ 1 + 2 + 3 + ... + k + (k + 1) <

1

8

[2(k+1)+1]2
Thus P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true ∀ n ∈ N.