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for this question
Prove the following statement by mathematical induction:
For all integers n >= 0, 2n < (n+2)!
is this answer right?
We need to follow two steps,
1. Base Case - we will prove that the statement holds for the smallest value in the domain which in this case n >= 0
2. Inductive Step Assume that the statement holds for some arbitrary positive integer k (the induction hypothesis) and then prove that it also holds for k + 1.
Base Case, n = 0.
2n = 2 * 0 = 0
(n + 2)! = (0 + 2)! = 2!
Since 0 < 2! is true, the base case holds!
Inductive Step, assume the statement is true for k where 2k < (k + 2)!, now prove the same for k + 1
Hypothesis: assume 2k < (k + 2)!
Now consider n = k + 1
2n = 2(k + 1) = 2k + 2
(n + 2)! = (k + 3)! = (k + 2)! * (k + 3)
Since we know 2k < (k + 2)! from the hypothesis we can write: 2k + 2 < (k + 2)! + 2
After simplifying we get: 2k < (k + 2)! which equals our hypothesis.
Therefore 2n < (n + 2)! holds!