Question:
2n < (n + 2)! for all natural number n.
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.