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

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- Bee_see

3^n < n!*

- zzr0ck3r

\(k>6\implies 3

- Bee_see

where did you get k> 6 from?

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## More answers

- zzr0ck3r

same thing as \(k\ge 7\)

- Bee_see

how?

- 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

\(k\ge7\implies 3

- zzr0ck3r

\((k+1)!=(k+1)k!\) always

- Bee_see

I have 3^k + (k+1) < k! + (k+1) thiough

- 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

multiplied what?

- Bee_see

where does the 3 < k+1 and the rest come from?

- freckles

3=7

- Bee_see

I don't have 3 < k+1 anywhere...

- 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

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

it isn't 3^k(3) < (k+1)!?

- Bee_see

I'm not sure from where you are starting and ending.

- freckles

? are you asking about my application of law of exponents ?

- freckles

\[x^{a+b}=x^{a} \cdot x^{b} \]
is the first thing I used

- freckles

\[\text{ we wanted \to show } 3^{k+1} <(k+1)!\]

- freckles

I started with the left hand side

- freckles

\[3^{k+1}=3^k 3^{1}=3^k (3)\]

- freckles

now using the inductive step we have:
\[3^k(3)=7 that 3 is going to be less than (k+1)

- Bee_see

so you are going to make 3^k+1 < (k+1)! into 3^k (k+1) < k! +(k+1?

- freckles

What are you doing ?

- freckles

where does all of that come from?

- freckles

we just wanted to show 3^(k+1) < (k+1)!

- Bee_see

ok..how did you make (k+1)! into k!(3)?

- freckles

do you understand that we have 3

- freckles

32
and we have k>=7 so this inequality 3

- freckles

k!*(k+1) is equal to (k+1)!

- freckles

just like 6!*7 is equal to 7!

- Bee_see

ok

- 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

just in one line like*

- Bee_see

ok

- freckles

I think I see what you were trying to do

- freckles

instead of adding (k+1) on both sides
it probably might be better if you multiply both sides by (k+1)

- freckles

\[3^{k}

- 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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