Let the given statement be P(n), i.e.,
P (n) : 12+32+52 + ... + (2n−1)2 =

n(2n−1)(2n+1)

3

First we prove that the statement is true for n = 1.
P (1) : 12 =

1.(2.1−1)(2.1+1)

3

⇒ 1 =

1.1.3

3

= 1 which is true
Assume P(k) is true for some positive integer k, i.e.,
12+32+52 + ... + (2k−1)2 =

k(2k−1)(2k+1)

3

... (i)
Now we shall prove that P(k + 1) is also true.
For this we have to prove that
12+32+52 + ... + (2k−1)2+(2(k+1)−1)2 =

(k+1)(2(k+1)−1)(2(k+1)+1)

3

L.H.S. = 12+32+52 + ... + (2k−1)2+(2(k+1)−1)2
=

k(2k−1)(2k+1)

3

+ [2(k+1)−1]2 [From (i)]
=

k(2k−1)(2k+1)

3

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

k(2k−1)

3

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

2k2−k+6k+3

3

] = (2k + 1) (

2k2+5k+3

3

)
=

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

3

=

(k+1)(2(k+1)−1)(2(k+1)+1)

3

= 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.