2n < ( n + 2 )! for all natural number n. Prove by mathematical induction
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Predicate statement P : 2 n < (n+2)!
For n = 1, then 2 n = 2 and (n+2)! = 3! = 6
So P is true.
Let us say P is true for n = k.
So 2 k < (k+2)!
2 k (k+3) < (k+2)! (k+3) = (k+3)!
6 k + 2k^2 < (k+3)!
2 (k+1) + 2(k^2+ 2k - 1) < (k+3)!
As k^2 + 2k - 1 is always positive for any k >= 1,
2 (k+1) < (k+3)! or (k+1 +2)!
Hence P is true for all natural numbers n.
For n = 1, then 2 n = 2 and (n+2)! = 3! = 6
So P is true.
Let us say P is true for n = k.
So 2 k < (k+2)!
2 k (k+3) < (k+2)! (k+3) = (k+3)!
6 k + 2k^2 < (k+3)!
2 (k+1) + 2(k^2+ 2k - 1) < (k+3)!
As k^2 + 2k - 1 is always positive for any k >= 1,
2 (k+1) < (k+3)! or (k+1 +2)!
Hence P is true for all natural numbers n.
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