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ashna

Prove that P(n) : n(n+1)(n+5) is divisible by 3

  • one year ago
  • one year ago

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  1. ashna
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    @waterineyes

    • one year ago
  2. joemath314159
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    What are our tool? Are you trying to do this with induction? or do we have modular arithmetic at our disposal?

    • one year ago
  3. ashna
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    induction

    • one year ago
  4. beginnersmind
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    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.

    • one year ago
  5. ashna
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    @beginnersmind i didn't get it :/ sorry can u explain it step by step ?

    • one year ago
  6. waterineyes
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    According to induction: Put n = 1 and see are you getting what the question says..

    • one year ago
  7. waterineyes
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    @beginnersmind do not stop. Carry on.. I am just trying as @ashna is doing..

    • one year ago
  8. ashna
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    i know that water tell me from , to prove p(k+1) is true

    • one year ago
  9. beginnersmind
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    My answer didn't use induction so I'd rather not go into a long explanation.

    • one year ago
  10. waterineyes
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    Ha ha ha ha... You knew that?? Just kidding..

    • one year ago
  11. ashna
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    we get k = 3M/ (K+1)(K+5) Right ?

    • one year ago
  12. waterineyes
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    Replace n by k+1 first...

    • one year ago
  13. ashna
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    yeah did , then ?

    • one year ago
  14. waterineyes
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    Then Look carefully it will also be divisible by 3.. Ha ha ha ha...

    • one year ago
  15. ashna
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    c'mon Water i don't understand :I

    • one year ago
  16. waterineyes
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    \[= (k+1)(k+2) (k+6)\]

    • one year ago
  17. ashna
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    okay

    • one year ago
  18. waterineyes
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    Really??

    • one year ago
  19. ashna
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    where r yu goin to substitute k = 3M/ (K+1)(K+5) ?

    • one year ago
  20. waterineyes
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    Wait...

    • one year ago
  21. ashna
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    okay

    • one year ago
  22. waterineyes
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    It is now 6 when I studied Induction..

    • one year ago
  23. waterineyes
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    *6 years..

    • one year ago
  24. beginnersmind
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    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.

    • one year ago
  25. ashna
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    yeah .. on assuming i got k = 3M/ (K+1)(K+5) 3rd step am stuck :I

    • one year ago
  26. beginnersmind
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    what does the M stand for?

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

    • one year ago
  28. ashna
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    M = divisible by 3

    • one year ago
  29. beginnersmind
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    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?

    • one year ago
  30. ashna
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    yes :)

    • one year ago
  31. beginnersmind
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    Cool :) To check, what did you get for P(k+1) - P(k) ?

    • one year ago
  32. waterineyes
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    What if we find the value of k+1 from the assumption??

    • one year ago
  33. waterineyes
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    \[k+1 = \frac{3M}{k (k+5)}\]

    • one year ago
  34. ashna
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    |dw:1355641516686:dw|

    • one year ago
  35. waterineyes
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    Hey I don't know how to prove that, I am just giving my Ideas which can be useless too..

    • one year ago
  36. waterineyes
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    Now this does not look like if it is divisible by 3 or not??

    • one year ago
  37. ashna
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    yeah .. :P

    • one year ago
  38. waterineyes
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    Wait, do not trust me.. Please confirm it from good source, may be I am wrong because I am not sure..

    • one year ago
  39. ashna
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    @beginnersmind suggest :)

    • one year ago
  40. beginnersmind
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    I don't think we are going in the right direction :( Let's just start by checking if the statement holds for n = 1

    • one year ago
  41. ashna
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    okay !

    • one year ago
  42. beginnersmind
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    And by let us, I mean you ;)

    • one year ago
  43. waterineyes
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    @UnkleRhaukus help here please..

    • one year ago
  44. beginnersmind
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    Let me rephrase: What's the value of P(1)?

    • one year ago
  45. ashna
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    12

    • one year ago
  46. ashna
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    divisible by 3 , so true !

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

    • one year ago
  48. ashna
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    p(k) = 3M / (k+1)(k+5)

    • one year ago
  49. beginnersmind
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    Let's forget about the induction hypothesis, just express P(k) and P(k+1) using the original definition

    • one year ago
  50. waterineyes
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    She has assumed that p(k) is divisible by 3 so 3M where M is an Integer..

    • one year ago
  51. ashna
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    i dont know how to do other than that :I @beginnersmind

    • one year ago
  52. beginnersmind
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    I mean, the same way you plugged in n = 1 to get P(1) you can plug in k to get P(k)

    • one year ago
  53. waterineyes
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    Just replace n by k there..

    • one year ago
  54. ashna
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    okay so k(k+1)(k=5) = divisible by 3 yu meant this ?

    • one year ago
  55. waterineyes
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    p(k) = .......................

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

    • one year ago
  57. ashna
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    okay , now how to proceed ?

    • one year ago
  58. beginnersmind
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    So the same way, what is P(k+1)?

    • one year ago
  59. ashna
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    P(k+1) = (k+1)(k+2)(k+6)

    • one year ago
  60. beginnersmind
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    right

    • one year ago
  61. ashna
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    and now ?

    • one year ago
  62. beginnersmind
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    We want to look at P(k+1) - P(k)

    • one year ago
  63. ashna
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    why ?

    • one year ago
  64. beginnersmind
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    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)

    • one year ago
  65. ashna
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    okay

    • one year ago
  66. beginnersmind
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    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?

    • one year ago
  67. ashna
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    yes i do

    • one year ago
  68. beginnersmind
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    Let's try to write out P(k+1) - P(k) and see if it's divisible by 3

    • one year ago
  69. beginnersmind
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    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.

    • one year ago
  70. waterineyes
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    It p(1) = 6 p(2) = 8 Then : p(2) - p(1) = 2

    • one year ago
  71. ashna
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    okay i've got (k+1)[(3k+12)]

    • one year ago
  72. waterineyes
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    This is divisible by 3 now...

    • one year ago
  73. beginnersmind
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    That's nice. I'd rewrite it as 3(k+1)(k+4) just to emphasize that it's divisible by 3.

    • one year ago
  74. ashna
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    right YAY :D ty @beginnersmind and @waterineyes

    • one year ago
  75. beginnersmind
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    Not finished yet. But almost :)

    • one year ago
  76. ashna
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    i can do the rest ty :)

    • one year ago
  77. beginnersmind
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    Ok :)

    • one year ago
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