2n < (n + 2)! for all natural

Solution:

According to the question,

P(n) is 2n < (n + 2)!

So, substituting different values for n, we get,

P(0) ⇒ 0 < 2!

P(1) ⇒ 2 < 3!

P(2) ⇒ 4 < 4!

P(3) ⇒ 6 < 5!

Let P(k) = 2k < (k + 2)! is true;

Now, we get that,

⇒ P(k+1) = 2(k+1) ((k+1)+2))!

We know that,

[(k+1)+2)! = (k+3)! = (k+3)(k+2)(k+1)……………3×2×1]

But, we also know that,

= 2(k+1) × (k+3)(k+2)……………3×1 > 2(k+1)

Therefore, 2(k+1) < ((k+1) + 2)!

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 2n < (n + 2)! Is true for all natural number n.