Prove the statement by the Principle of Mathematical Induction :n2 <

1.	Prove the statement by the Principle of Mathematical Induction :n2 < 2n for all natural numbers n ≥ 5.
Answer» 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.

Prove the statement by the Principle of Mathematical Induction :n2 < 2n for all natural numbers n ≥ 5.

Answer»

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.