Question

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

Answer 1

Let P(n) : n² < 2ⁿ for all natural numbers, n ≥ 5 

Step 1 : 

P(5) : 1⁵ < 2⁵ ⇒ 1 < 32 which is true for P(5) 

Step 2 :

P(k): k² < 2k. Let it be true for k ∈ N 

Step 3 : 

P(k + 1): (k + 1)² < 2^{k + 1}

From Step 2, we get k² < 2^k 

⇒ k² < 2^{k + 1} < 2^k + 2^{k + 1}

From eqn. (i) and (ii), we get (k + 1)² < 2^{k + 1} 

Hence, P(k + 1) is true whenever P(k) is true for k ∈ N, n ≥ 5.