## ashna Group Title Prove that P(n) : n(n+1)(n+5) is divisible by 3 one year ago one year ago

1. ashna Group Title

@waterineyes

2. joemath314159 Group Title

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

3. ashna Group Title

induction

4. beginnersmind Group Title

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. ashna Group Title

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

6. waterineyes Group Title

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

7. waterineyes Group Title

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

8. ashna Group Title

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

9. beginnersmind Group Title

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

10. waterineyes Group Title

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

11. ashna Group Title

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

12. waterineyes Group Title

Replace n by k+1 first...

13. ashna Group Title

yeah did , then ?

14. waterineyes Group Title

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

15. ashna Group Title

c'mon Water i don't understand :I

16. waterineyes Group Title

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

17. ashna Group Title

okay

18. waterineyes Group Title

Really??

19. ashna Group Title

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

20. waterineyes Group Title

Wait...

21. ashna Group Title

okay

22. waterineyes Group Title

It is now 6 when I studied Induction..

23. waterineyes Group Title

*6 years..

24. beginnersmind Group Title

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. ashna Group Title

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

26. beginnersmind Group Title

what does the M stand for?

27. beginnersmind Group Title

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

28. ashna Group Title

M = divisible by 3

29. beginnersmind Group Title

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. ashna Group Title

yes :)

31. beginnersmind Group Title

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

32. waterineyes Group Title

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

33. waterineyes Group Title

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

34. ashna Group Title

|dw:1355641516686:dw|

35. waterineyes Group Title

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

36. waterineyes Group Title

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

37. ashna Group Title

yeah .. :P

38. waterineyes Group Title

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

39. ashna Group Title

@beginnersmind suggest :)

40. beginnersmind Group Title

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. ashna Group Title

okay !

42. beginnersmind Group Title

And by let us, I mean you ;)

43. waterineyes Group Title

44. beginnersmind Group Title

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

45. ashna Group Title

12

46. ashna Group Title

divisible by 3 , so true !

47. beginnersmind Group Title

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

48. ashna Group Title

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

49. beginnersmind Group Title

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

50. waterineyes Group Title

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

51. ashna Group Title

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

52. beginnersmind Group Title

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

53. waterineyes Group Title

Just replace n by k there..

54. ashna Group Title

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

55. waterineyes Group Title

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

56. beginnersmind Group Title

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. ashna Group Title

okay , now how to proceed ?

58. beginnersmind Group Title

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

59. ashna Group Title

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

60. beginnersmind Group Title

right

61. ashna Group Title

and now ?

62. beginnersmind Group Title

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

63. ashna Group Title

why ?

64. beginnersmind Group Title

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. ashna Group Title

okay

66. beginnersmind Group Title

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. ashna Group Title

yes i do

68. beginnersmind Group Title

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

69. beginnersmind Group Title

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. waterineyes Group Title

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

71. ashna Group Title

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

72. waterineyes Group Title

This is divisible by 3 now...

73. beginnersmind Group Title

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

74. ashna Group Title

right YAY :D ty @beginnersmind and @waterineyes

75. beginnersmind Group Title

Not finished yet. But almost :)

76. ashna Group Title

i can do the rest ty :)

77. beginnersmind Group Title

Ok :)