n^2 < 2^n for all natural numbers n ≥ 5.

Question

Prove the statement by the Principle of Mathematical Induction :

n² < 2ⁿ for all natural numbers n ≥ 5.

Answer 1

According to the question,

P(n) is n² < 2ⁿ  for n≥5

Let P(k) = k² < 2^k be true;

⇒ P(k+1) = (k+1)²

= k² + 2k + 1

2^k+1 = 2(2^k) > 2k²

Since, n² > 2n + 1 for n ≥3

We get that,

k² + 2k + 1 < 2k²

⇒ (k+1)² < 2^(k+1)

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = n² < 2ⁿ is true for all natural numbers n ≥ 5.