**1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers .**

###### Answers:

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(l) is true.

Let us assume that P(n) is true for some natural number n = k.

∴ P(k):l+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]

= 2k^{2} -k+4k+ 4-3

= 2k^{2} + 3k + 1

= (k+ 1)( 2k + 1)

= (k+l)[2(k+l)-l]

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.