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Question
Prove the statement by using the Principle of Mathematical Induction:
2n < (n + 2)! for all natural number n.
Solution
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.
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