A community for students.
Here's the question you clicked on:
 0 viewing
Bee_see
 one year ago
Prove by induction that
3n < n!
when n is an integer and n >= 7.
I have proven that when n=7, the proof is true...
So I did n=k, 3^k < k! then
3^k+1 < (k+1)! I added the term k+1 on both sides...3^k+(k+1) < k! + (k+1). How do go to 3^k+1 < (k+1)! ?
Bee_see
 one year ago
Prove by induction that 3n < n! when n is an integer and n >= 7. I have proven that when n=7, the proof is true... So I did n=k, 3^k < k! then 3^k+1 < (k+1)! I added the term k+1 on both sides...3^k+(k+1) < k! + (k+1). How do go to 3^k+1 < (k+1)! ?

This Question is Closed

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3\(k>6\implies 3<k+1\implies 3*3^k<(k+1)3^k\overset{\text{IH}}{<}(k+1)k!\implies 3^{k+1}<(k+1)!\)

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0where did you get k> 6 from?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3same thing as \(k\ge 7\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3k is an integer. \(k>6\implies 7,8,9,10,11,12...\) \(k\ge 7 \implies 7,8,9,10,11,12...\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3\(k\ge7\implies 3<k+1\implies 3*3^k<(k+1)3^k\overset{\text{IH}}{<}(k+1)k!\implies 3^{k+1}<(k+1)!\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3\((k+1)!=(k+1)k!\) always

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0I have 3^k + (k+1) < k! + (k+1) thiough

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3You used addition to try and get there, I used multiplication. I am not saying you cant do it your way, but this is the only way I can do it.

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0where does the 3 < k+1 and the rest come from?

freckles
 one year ago
Best ResponseYou've already chosen the best response.13<k+1 for all integer k>=7

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0I don't have 3 < k+1 anywhere...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[3^{k+1}=3^k \cdot 3^1 \text{ by law of exponents } \\ 3^{k}3<k!(3) \text{ since }3^{k}<k! \\ k!(3)<k!(k+1) \text{ since for } k \ge 7 \text{ we have } k+1>3 \\ \text{ finally } k!(k+1)=(k+1)!\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1we do have k+1>3 though for k>=7 7+1>3 8+1>3 9+1>3 10+1>3 and so on...

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0it isn't 3^k(3) < (k+1)!?

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure from where you are starting and ending.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1? are you asking about my application of law of exponents ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[x^{a+b}=x^{a} \cdot x^{b} \] is the first thing I used

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\text{ we wanted \to show } 3^{k+1} <(k+1)!\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I started with the left hand side

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[3^{k+1}=3^k 3^{1}=3^k (3)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1now using the inductive step we have: \[3^k(3)<k!(3)\] now we want to show this is <(k+1)!=k!*(k+1) and we know since k>=7 that 3 is going to be less than (k+1)

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0so you are going to make 3^k+1 < (k+1)! into 3^k (k+1) < k! +(k+1?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1where does all of that come from?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1we just wanted to show 3^(k+1) < (k+1)!

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0ok..how did you make (k+1)! into k!(3)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1do you understand that we have 3<k+1?

freckles
 one year ago
Best ResponseYou've already chosen the best response.13<k+1 holds for any integer k>2 and we have k>=7 so this inequality 3<k+1 is true for our k

freckles
 one year ago
Best ResponseYou've already chosen the best response.1k!*(k+1) is equal to (k+1)!

freckles
 one year ago
Best ResponseYou've already chosen the best response.1just like 6!*7 is equal to 7!

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[3^{k+1}=3^k \cdot 3^1 \text{ by law of exponents } \\ 3^{k}3<k!(3) \text{ since }3^{k}<k! \\ k!(3)<k!(k+1) \text{ since for } k \ge 7 \text{ we have } k+1>3 \\ \text{ finally } k!(k+1)=(k+1)!\] I wrote this like this to show my reasons for each steps you can write this without all the reasons...you know just in one like: \[3^{k+1}=3^{k} \cdot 3^1 =3^{k} 3<k! (3)<k! \cdot (k+1)=(k+1)!\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1just in one line like*

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I think I see what you were trying to do

freckles
 one year ago
Best ResponseYou've already chosen the best response.1instead of adding (k+1) on both sides it probably might be better if you multiply both sides by (k+1)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[3^{k}<k! \\ \text{ multiply both sides by } (k+1) \\ 3^k(k+1)<k! \cdot (k+1) \\ \text{ recall } k!(k+1)=(k+1)! \\ \text{ so we have } 3^k(k+1) <(k+1)! \\ \text{ but we still kind of need } 3<k+1 \text{ which is true since } k \ge 7 \\ 3^{k}(3)<3^k(k+1)<k!(k+1)=(k+1)! \\ \text{ and we know by law of exponents } 3^k(3)=3^{k+1} \\ \text{ so add this \to our little line } \\ 3^{k+1}=3^k(3)<3^k(k+1)<k!(k+1)=(k+1)!\] still the same line we had above

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I just decided to work from the side that had 3^(k+1) to show this was less than (k+1)!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.