Prove the statement the Principle of Mathematical Induction:1 + 5 + 9

1.	Prove the statement the Principle of Mathematical Induction:1 + 5 + 9 + ... + (4n – 3) = n (2n – 1) for all natural numbers n.
Answer» 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.

Prove the statement the Principle of Mathematical Induction:1 + 5 + 9 + ... + (4n – 3) = n (2n – 1) for all natural numbers n.

Answer»

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.