Let P(n): 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 x 1 – 1) = 1, which is true.
Hence, P(1) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k) :1 + 5 + 9 +…+(4k -3) = k(2k - 1) .......(i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1) : 1 + 5 + 9 + … + (4k - 3) + [4(k + 1) – 3]
= 2k2 - k + 4k + 4 - 3
= 2k2 + 3k + 1
= (k + 1)( 2k + 1)
= (k + 1)[2(k + 1) - 1]
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.