## 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)! ?

1. Bee_see

3^n < n!*

2. zzr0ck3r

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

3. Bee_see

where did you get k> 6 from?

4. zzr0ck3r

same thing as $$k\ge 7$$

5. Bee_see

how?

6. zzr0ck3r

k is an integer. $$k>6\implies 7,8,9,10,11,12...$$ $$k\ge 7 \implies 7,8,9,10,11,12...$$

7. zzr0ck3r

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

8. zzr0ck3r

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

9. Bee_see

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

10. 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.

11. Bee_see

multiplied what?

12. Bee_see

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

13. freckles

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

14. Bee_see

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

15. freckles

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

16. 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...

17. Bee_see

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

18. Bee_see

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

19. freckles

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

20. freckles

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

21. freckles

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

22. freckles

I started with the left hand side

23. freckles

$3^{k+1}=3^k 3^{1}=3^k (3)$

24. freckles

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)

25. Bee_see

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

26. freckles

What are you doing ?

27. freckles

where does all of that come from?

28. freckles

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

29. Bee_see

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

30. freckles

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

31. freckles

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

32. freckles

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

33. freckles

just like 6!*7 is equal to 7!

34. Bee_see

ok

35. freckles

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

36. freckles

just in one line like*

37. Bee_see

ok

38. freckles

I think I see what you were trying to do

39. freckles

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

40. freckles

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

41. freckles

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