Mathematica Mathematical Induction trouble with 1 step

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The discussion revolves around proving the inequality n! > 4^n for all n ≥ 9 using mathematical induction. The user successfully verifies the base case for n = 9 and assumes the induction hypothesis for n = k. The challenge arises when trying to prove the case for n = k + 1, specifically how to show that (k + 1)(4^k) > 4^(k + 1). The user realizes that since k is at least 9, k + 1 will also be greater than 4, which clarifies the proof. Ultimately, the user acknowledges that minor mistakes hindered their progress but appreciates the insight gained.
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Hi there folks, I have just a small problem with a specific induction problem. The problem itself is: "Prove n! > 4^n, for all n >= 9."

So here's my work:

1) Show true for n = 9
LS
9! = 362880

RS
4^9 = 262144

.:. LS > RS

2) Assume true for n = k
i.e. Assume that k! > 4^k

3) Prove true for n = k+1
i.e. Prove that (k+1)! > 4^(k+1)

So I begin to expand the LHS out

(k+1)! = (k+1)(k)!
> (K+1)(4^k) (by induction hypothesis)this is the problem that I encounter. I get stuck here because I don't exactly know how to followthrough at this point. How does (k+1)(4^k) become greater that 4^(k+1) ?
 
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If k >= 9, then what can you say about k + 1? How is k + 1 compared to 4 (greater, less than, or equal to)?
So from there can you prove that (k + 1)4k > 4k + 1 = 4 4k?
 
VietDao29 said:
If k >= 9, then what can you say about k + 1? How is k + 1 compared to 4 (greater, less than, or equal to)?
So from there can you prove that (k + 1)4k > 4k + 1 = 4 4k?

OH. I see, since k always going to be at least 10. I didn't see it that way.

Dammit! Oh well, a few silly mistakes cost me quite alot.

Thanks.
 
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