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

###### Answers:

Let P(n): 2n < (n + 2)! for all natural numbers n.

P( 1): 2 < (1 + 2)! or 2 < 3! or 2 < 6, which is true.

Hence,P(l) is true.

Let us assume that P(n) is true for some natural number n = k.

P(k) :2k<(k + 2)! (i)

To prove that P(k + 1) is true, we have to show that

P(k + 1): 2(k+ 1) < (k + 1 + 2)!

or 2(k+ 1) < (k + 3)!

Using (i), we get

2(k + 1) = 2k + 2<(k+2) ! +2 (ii)

Now let, (k + 2)! + 2 < (k + 3)! (iii)

⇒ 2 < (k+ 3)! – (k+2) !

⇒ 2 < (k + 2) ! [k+ 3-1]

⇒ 2<(k+ 2) ! (k + 2), which is true for any natural number.

Using (ii) and (iii), we get 2(k + 1) < (k + 3)!

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.