Question

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)! ?

Answer 1

3^n < n!*

Answer 2

\(k>6\implies 3<k+1\implies 3*3^k<(k+1)3^k\overset{\text{IH}}{<}(k+1)k!\implies 3^{k+1}<(k+1)!\)

Answer 3

where did you get k> 6 from?

Answer 4

same thing as \(k\ge 7\)

Answer 5

how?

Answer 6

k is an integer.

\(k>6\implies 7,8,9,10,11,12...\)


\(k\ge 7 \implies 7,8,9,10,11,12...\)

Answer 7

\(k\ge7\implies 3<k+1\implies 3*3^k<(k+1)3^k\overset{\text{IH}}{<}(k+1)k!\implies 3^{k+1}<(k+1)!\)

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

multiplied what?

Answer 12

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

Answer 13

3<k+1 for all integer k>=7

Answer 14

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

Answer 15

\[3^{k+1}=3^k \cdot 3^1 \text{ by law of exponents } \\ 3^{k}3<k!(3) \text{ since }3^{k}<k! \\ k!(3)<k!(k+1) \text{ since for } k \ge 7 \text{ we have } k+1>3 \\ \text{ finally } k!(k+1)=(k+1)!\]

Answer 16

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...

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

I started with the left hand side

Answer 23

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

Answer 24

now 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)

Answer 25

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

Answer 26

What are you doing ?

Answer 27

where does all of that come from?

Answer 28

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

Answer 29

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

Answer 30

do you understand that we have 3<k+1?

Answer 31

3<k+1 holds for any integer k>2
and we have k>=7 so this inequality 3<k+1 is true for our k

Answer 32

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

Answer 33

just like 6!*7 is equal to 7!

Answer 34

ok

Answer 35

\[3^{k+1}=3^k \cdot 3^1 \text{ by law of exponents } \\ 3^{k}3<k!(3) \text{ since }3^{k}<k! \\ k!(3)<k!(k+1) \text{ since for } k \ge 7 \text{ we have } k+1>3 \\ \text{ finally } k!(k+1)=(k+1)!\]
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:
\[3^{k+1}=3^{k} \cdot 3^1 =3^{k} 3<k! (3)<k! \cdot (k+1)=(k+1)!\]

Answer 36

just in one line like*

Answer 37

ok

Answer 38

I think I see what you were trying to do

Answer 39

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

Answer 40

\[3^{k}<k! \\ \text{ multiply both sides by } (k+1) \\ 3^k(k+1)<k! \cdot (k+1) \\ \text{ recall } k!(k+1)=(k+1)! \\ \text{ so we have } 3^k(k+1) <(k+1)! \\ \text{ but we still kind of need } 3<k+1 \text{ which is true since } k \ge 7 \\ 3^{k}(3)<3^k(k+1)<k!(k+1)=(k+1)! \\ \text{ and we know by law of exponents } 3^k(3)=3^{k+1} \\ \text{ so add this \to our little line } \\ 3^{k+1}=3^k(3)<3^k(k+1)<k!(k+1)=(k+1)!\]
still the same line we had above

Answer 41

I just decided to work from the side that had 3^(k+1) to show this was less than (k+1)!