## anonymous 3 years ago Prove that P(n) : n(n+1)(n+5) is divisible by 3

1. anonymous

@waterineyes

2. anonymous

What are our tool? Are you trying to do this with induction? or do we have modular arithmetic at our disposal?

3. anonymous

induction

4. beginnersmind

My favourite would be to write it as n(n+1)(n+2) + 3n(n+1) The second term is divisible by 3 and the first one is the product of 3 consecutive integers therefore also divisible by 3. The sum of 2 numbers both divisible by 3 is also divisible by 3.

5. anonymous

@beginnersmind i didn't get it :/ sorry can u explain it step by step ?

6. anonymous

According to induction: Put n = 1 and see are you getting what the question says..

7. anonymous

@beginnersmind do not stop. Carry on.. I am just trying as @ashna is doing..

8. anonymous

i know that water tell me from , to prove p(k+1) is true

9. beginnersmind

My answer didn't use induction so I'd rather not go into a long explanation.

10. anonymous

Ha ha ha ha... You knew that?? Just kidding..

11. anonymous

we get k = 3M/ (K+1)(K+5) Right ?

12. anonymous

Replace n by k+1 first...

13. anonymous

yeah did , then ?

14. anonymous

Then Look carefully it will also be divisible by 3.. Ha ha ha ha...

15. anonymous

c'mon Water i don't understand :I

16. anonymous

$= (k+1)(k+2) (k+6)$

17. anonymous

okay

18. anonymous

Really??

19. anonymous

where r yu goin to substitute k = 3M/ (K+1)(K+5) ?

20. anonymous

Wait...

21. anonymous

okay

22. anonymous

It is now 6 when I studied Induction..

23. anonymous

*6 years..

24. beginnersmind

Ok, this is how you do it with induction. First prove it for n =1 (plug it in and check if it's divisible by 3) Second assume that it's true for P(k). Using this try to prove it's also true for P(k+1) In this case I would try to prove that P(k+1) - P(k) is divisible by 3.

25. anonymous

yeah .. on assuming i got k = 3M/ (K+1)(K+5) 3rd step am stuck :I

26. beginnersmind

what does the M stand for?

27. beginnersmind

Ah, ok, see what you did there. You said there's a number M such that P(k) = 3M

28. anonymous

M = divisible by 3

29. beginnersmind

Ok, I'd do it slightly differently. I'd prove that the difference of P(k+1) and P(k) is divisible by 3. Then using this and the induction hypothesis it follows that P(k+1) is also divisible by 3. Does that make sense?

30. anonymous

yes :)

31. beginnersmind

Cool :) To check, what did you get for P(k+1) - P(k) ?

32. anonymous

What if we find the value of k+1 from the assumption??

33. anonymous

$k+1 = \frac{3M}{k (k+5)}$

34. anonymous

|dw:1355641516686:dw|

35. anonymous

Hey I don't know how to prove that, I am just giving my Ideas which can be useless too..

36. anonymous

Now this does not look like if it is divisible by 3 or not??

37. anonymous

yeah .. :P

38. anonymous

Wait, do not trust me.. Please confirm it from good source, may be I am wrong because I am not sure..

39. anonymous

@beginnersmind suggest :)

40. beginnersmind

I don't think we are going in the right direction :( Let's just start by checking if the statement holds for n = 1

41. anonymous

okay !

42. beginnersmind

And by let us, I mean you ;)

43. anonymous

44. beginnersmind

Let me rephrase: What's the value of P(1)?

45. anonymous

12

46. anonymous

divisible by 3 , so true !

47. beginnersmind

Ok. So we've proven that it holds for n =1. Now, what's P(k)? P(k+1)?

48. anonymous

p(k) = 3M / (k+1)(k+5)

49. beginnersmind

Let's forget about the induction hypothesis, just express P(k) and P(k+1) using the original definition

50. anonymous

She has assumed that p(k) is divisible by 3 so 3M where M is an Integer..

51. anonymous

i dont know how to do other than that :I @beginnersmind

52. beginnersmind

I mean, the same way you plugged in n = 1 to get P(1) you can plug in k to get P(k)

53. anonymous

Just replace n by k there..

54. anonymous

okay so k(k+1)(k=5) = divisible by 3 yu meant this ?

55. anonymous

p(k) = .......................

56. beginnersmind

I mean P(k) = k(k+1)(k+5) We'll use the fact that it's divisible by 3 later, but not yet.

57. anonymous

okay , now how to proceed ?

58. beginnersmind

So the same way, what is P(k+1)?

59. anonymous

P(k+1) = (k+1)(k+2)(k+6)

60. beginnersmind

right

61. anonymous

and now ?

62. beginnersmind

We want to look at P(k+1) - P(k)

63. anonymous

why ?

64. beginnersmind

This is the difference between two consecutive terms. For example the difference between the second term and the first term is P(2)-P(1)

65. anonymous

okay

66. beginnersmind

If P(2) - P(1) is divisible by 3 and P(1) is divisible by 3 then so is P(2). Do you see why?

67. anonymous

yes i do

68. beginnersmind

Let's try to write out P(k+1) - P(k) and see if it's divisible by 3

69. beginnersmind

You already calculated P(k+1) and P(k) separately so just take the difference and see if you can write it in some nice way.

70. anonymous

It p(1) = 6 p(2) = 8 Then : p(2) - p(1) = 2

71. anonymous

okay i've got (k+1)[(3k+12)]

72. anonymous

This is divisible by 3 now...

73. beginnersmind

That's nice. I'd rewrite it as 3(k+1)(k+4) just to emphasize that it's divisible by 3.

74. anonymous

right YAY :D ty @beginnersmind and @waterineyes

75. beginnersmind

Not finished yet. But almost :)

76. anonymous

i can do the rest ty :)

77. beginnersmind

Ok :)