Let the given statement be P(n), i.e.,
P (n) : 1 . 3 + 3 . 5 + 5 . 7 + ... + (2n - 1) (2n + 1) =

n(4n2+6n−1)

3

First we prove that, the statement is true for n = 1.
P (1) : 1 . 3 =

1.(4.(1)2+6.1−1)

3

⇒ 3 =

4+6−1

3

=

9

3

= 3 , which is true
Assume P(k) is true for some positive integer k, i.e.,
1 . 3 + 3 . 5 + 5 . 7 + ... + (2k - 1) (2k + 1) =

k(4k2+6k−1)

3

... (i)
We shall now prove that P(k + 1) is also true.
For this we have to prove that
1·3 + 3·5 + 5·7 + ........ + (2k – 1)(2k + 1) + (2k + 1)(2k + 3)
=

(k+1)[4(k+1)2+6(k+1)−1]

3

L.H.S. = 1·3 + 3·5 + 5·7 + ..... + (2k – 1)(2k + 1) + (2k + 1)(2k + 3)
=

k(4k2+6k−1)

3

+ (2k + 1) (2k + 3) [From (i)]
=

4k3+6k2−k+3(4k2+8k+3)

3

=

4k3+6k2−k+12k2+24k+9

3

=

4k3+18k2+23k+9

3

... (ii)
Also, R.H.S. =

(k+1)[4(k+1)2+6(k+1)−1]

3

= =

(k+1)[4k2+8k+4+6k+5]

3

=

(k+1)[4k2+14k+9]

3

=

4k3+14k2+9k+4k2+14k+9

3

=

4k3+18k2+23k+9

3

... (iii)
From (ii) and (iii), we get L.H.S. = 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.