n2 < 2n for all natural numbers

According to the question,

P(n) is n2 < 2n  for n≥5

Let P(k) = k2 < 2k be true;

⇒ P(k+1) = (k+1)2

= k2 + 2k + 1

2k+1 = 2(2k) > 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.