Bee_see
  • Bee_see
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)! ?
Mathematics
katieb
  • katieb
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Bee_see
  • Bee_see
3^n < n!*
zzr0ck3r
  • zzr0ck3r
\(k>6\implies 3
Bee_see
  • Bee_see
where did you get k> 6 from?

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zzr0ck3r
  • zzr0ck3r
same thing as \(k\ge 7\)
Bee_see
  • Bee_see
how?
zzr0ck3r
  • zzr0ck3r
k is an integer. \(k>6\implies 7,8,9,10,11,12...\) \(k\ge 7 \implies 7,8,9,10,11,12...\)
zzr0ck3r
  • zzr0ck3r
\(k\ge7\implies 3
zzr0ck3r
  • zzr0ck3r
\((k+1)!=(k+1)k!\) always
Bee_see
  • Bee_see
I have 3^k + (k+1) < k! + (k+1) thiough
zzr0ck3r
  • zzr0ck3r
You 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
  • Bee_see
multiplied what?
Bee_see
  • Bee_see
where does the 3 < k+1 and the rest come from?
freckles
  • freckles
3=7
Bee_see
  • Bee_see
I don't have 3 < k+1 anywhere...
freckles
  • freckles
\[3^{k+1}=3^k \cdot 3^1 \text{ by law of exponents } \\ 3^{k}33 \\ \text{ finally } k!(k+1)=(k+1)!\]
freckles
  • freckles
we 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
  • Bee_see
it isn't 3^k(3) < (k+1)!?
Bee_see
  • Bee_see
I'm not sure from where you are starting and ending.
freckles
  • freckles
? are you asking about my application of law of exponents ?
freckles
  • freckles
\[x^{a+b}=x^{a} \cdot x^{b} \] is the first thing I used
freckles
  • freckles
\[\text{ we wanted \to show } 3^{k+1} <(k+1)!\]
freckles
  • freckles
I started with the left hand side
freckles
  • freckles
\[3^{k+1}=3^k 3^{1}=3^k (3)\]
freckles
  • freckles
now using the inductive step we have: \[3^k(3)=7 that 3 is going to be less than (k+1)
Bee_see
  • Bee_see
so you are going to make 3^k+1 < (k+1)! into 3^k (k+1) < k! +(k+1?
freckles
  • freckles
What are you doing ?
freckles
  • freckles
where does all of that come from?
freckles
  • freckles
we just wanted to show 3^(k+1) < (k+1)!
Bee_see
  • Bee_see
ok..how did you make (k+1)! into k!(3)?
freckles
  • freckles
do you understand that we have 3
freckles
  • freckles
32 and we have k>=7 so this inequality 3
freckles
  • freckles
k!*(k+1) is equal to (k+1)!
freckles
  • freckles
just like 6!*7 is equal to 7!
Bee_see
  • Bee_see
ok
freckles
  • freckles
\[3^{k+1}=3^k \cdot 3^1 \text{ by law of exponents } \\ 3^{k}33 \\ \text{ finally } k!(k+1)=(k+1)!\] 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
freckles
  • freckles
just in one line like*
Bee_see
  • Bee_see
ok
freckles
  • freckles
I think I see what you were trying to do
freckles
  • freckles
instead of adding (k+1) on both sides it probably might be better if you multiply both sides by (k+1)
freckles
  • freckles
\[3^{k}
freckles
  • freckles
I just decided to work from the side that had 3^(k+1) to show this was less than (k+1)!

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