prove that `2nlt(n+2)!` for all natural numbers n.

Explanation

Consider the statement
`P(n):2nlt(n+2)!` for all natural number n.
Step I We observe that, P(1) is true `P(1):2(1)lt(1+2)!`
`rArr 2lt3!rArr2lt3xx2xx1rArr2lt6`
Hence, P(1) is true.
Step II Now, assume that p(n) is true for n=k,
`P(k):2klt(k+2)!` is true
Step III Now, assume that P(n) is true for n=k,
`P(k):2klt(k+2)!` is true.
Step III To prove P(k+1) is true, we have to show that
`P(k+1):2(k+1)lt(k+1+2)!`
Now, `2klt(k+2)!`
`2k+2lt(k+2)!`
`2k+2ltIk+2)!+2`
`2(k+1)lt(k+2)!+2` . . .(i)
Also, `(k+2)!+2lt(k+3)!` . . . (ii)
From Eqs. (i) and (ii),
`2(k+1)lt(k+1+2)!`
So, P(k+1) is true, whenever P(k) is true.
Hence, by principle of mathematical induction P(n) is true.