1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = n(4n2 + 6n – 1)/3

Let us check for n = 1,

P (1): (2.1 – 1) (2.1 + 1) = \(\dfrac{ 1(4.1^2 + 6.1 -1)}{3}\)

1×3 = \(\dfrac{ 1(4+6-1)}{3}\)

3 = 3

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) =\( \dfrac{ k(4k^2 + 6k – 1)}{3}\) … (i)

So,

1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)

Now, substituting the value of P (k) we get,

= \(\dfrac{ k(4k^2 + 6k – 1)}{3}\) + (2k + 1) (2k + 3) by using equation (i)

=\( \dfrac{(k+1) (4k^2 + 8k +4 + 6k + 6 – 1)] }{ 3}\)

=\( \dfrac{(k+1) 4(k+1)2 + 6(k+1) -1 }{3}\)

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.