Use induction to prove that n! > 3n for all n > 6

[solomon_13000]

Use induction to prove that n! > 3n for all n > 6.

My work:

Assume n = p
the next step n = p + 1

(p+1)! > 3(p+1)

(p+1)! = (p)! x (p+1)
> 3p x (p+1)

how did (p+1)! expand into (p)! x (p+1)
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Edited by stapel -- Reason for edit: clarification, standard English, etc

 

[stapel]

Are you going to show the "n = 6" step? (This is, as has been explained a few times previously, the necessary first step.)

Note: The "n = p" step is not "Let's now assume that n = p, and then do nothing with that." As has been explained, the "n = p" step is where you make your assumption. What is the assumption in this case? (You need to state this clearly as part of your proof.)

For the "n = p + 1" step, you aren't trying to "solve" the equation or inequality. As previously mentioned (in other threads), you are trying to manipulate one side of the "n = p + 1" step to match or fit the other side. From what you've posted (it's a bit cryptic), I will guess that you are working with the left-hand side:

. . . . .(p + 1)!

What does (p + 1)! equal? In particular, how can you incorporate your assumption from the "n = p" step?

how did (p+1)! expand into (p)! x (p+1)

Are you not familiar with factorials...?

Thank you.

Eliz.

 

[solomon_13000]

ok now I understand the factorial part.

(n+1)! = (n)! x (n+1)