Stephanie M. answered 07/22/15
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For mathematical induction, you'll want to follow three steps:
1. PROVE THE STATEMENT IS TRUE FOR THE SMALLEST VALUE OF N
2. ASSUME THE STATEMENT IS TRUE FOR N (OR N - 1)
3. PROVE THE STATEMENT IS TRUE FOR N + 1 (OR N)
Let's do that for your proof.
STEP 1. Prove the statement is true for the smallest value of n.
For you, the smallest value allowed is n = 5. So:
(5+1)! = 6! = 6×5×4×3×2×1 = 720
35 = 3×3×3×3×3 = 243
Therefore, (5+1)! > 35.
STEP 2. Assume the statement is true for n (or n - 1).
We'll use n. So, assume the following is true:
(n + 1)! > 3n
STEP 3. Prove the statement is true for n + 1 (or n).
Since we used n in Step 2, we need to prove the statement is true for n + 1 here.
(n + 1 + 1)! = (n + 2)! = (n + 1)! × (n + 2)
3n+1 = 3n × 31 = 3n × 3
From our assumption, we know that (n + 1)! > 3n. We also know, since n > 4, that (n + 2) > 3. So, the product of (n + 1)! and (n + 2) must be greater than the product of 3n and 3, since the components of the first product are each greater than the corresponding components of the second product. In other words, if you start with a larger number and a smaller number, then multiply the larger number by another larger number and the smaller number by another smaller number, the first product is bigger than the second product.
So, (n + 1)! × (n + 2) > 3n × 3, which means that (n + 1 + 1)! > 3n+1.
Thus, we've proven the statement for the n + 1 case.
QED.
This concludes the proof. This method works because we've basically proven that (1) if the statement is true for a given case, we could prove that it was true for the next case. And we've proven that (2) the statement is true for the smallest case.
Given those two facts, we could conceivably start with the smallest case, then prove each successive case. In other words, we could prove the statement is true for all n > 4.