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ashna

  • 2 years ago

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

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  1. ashna
    • 2 years ago
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    @waterineyes

  2. joemath314159
    • 2 years ago
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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?

  3. ashna
    • 2 years ago
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    induction

  4. beginnersmind
    • 2 years ago
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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.

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

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

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

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

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

  10. waterineyes
    • 2 years ago
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    Ha ha ha ha... You knew that?? Just kidding..

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

  12. waterineyes
    • 2 years ago
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    Replace n by k+1 first...

  13. ashna
    • 2 years ago
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    yeah did , then ?

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

  15. ashna
    • 2 years ago
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    c'mon Water i don't understand :I

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

  17. ashna
    • 2 years ago
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    okay

  18. waterineyes
    • 2 years ago
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    Really??

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

  20. waterineyes
    • 2 years ago
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    Wait...

  21. ashna
    • 2 years ago
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    okay

  22. waterineyes
    • 2 years ago
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    It is now 6 when I studied Induction..

  23. waterineyes
    • 2 years ago
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    *6 years..

  24. beginnersmind
    • 2 years ago
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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.

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

  26. beginnersmind
    • 2 years ago
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    what does the M stand for?

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

  28. ashna
    • 2 years ago
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    M = divisible by 3

  29. beginnersmind
    • 2 years ago
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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?

  30. ashna
    • 2 years ago
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    yes :)

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

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

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

  34. ashna
    • 2 years ago
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    |dw:1355641516686:dw|

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

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

  37. ashna
    • 2 years ago
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    yeah .. :P

  38. waterineyes
    • 2 years ago
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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..

  39. ashna
    • 2 years ago
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    @beginnersmind suggest :)

  40. beginnersmind
    • 2 years ago
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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

  41. ashna
    • 2 years ago
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    okay !

  42. beginnersmind
    • 2 years ago
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    And by let us, I mean you ;)

  43. waterineyes
    • 2 years ago
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    @UnkleRhaukus help here please..

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

  45. ashna
    • 2 years ago
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    12

  46. ashna
    • 2 years ago
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    divisible by 3 , so true !

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

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

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

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

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

  52. beginnersmind
    • 2 years ago
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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)

  53. waterineyes
    • 2 years ago
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    Just replace n by k there..

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

  55. waterineyes
    • 2 years ago
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    p(k) = .......................

  56. beginnersmind
    • 2 years ago
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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.

  57. ashna
    • 2 years ago
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    okay , now how to proceed ?

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

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

  60. beginnersmind
    • 2 years ago
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    right

  61. ashna
    • 2 years ago
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    and now ?

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

  63. ashna
    • 2 years ago
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    why ?

  64. beginnersmind
    • 2 years ago
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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)

  65. ashna
    • 2 years ago
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    okay

  66. beginnersmind
    • 2 years ago
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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?

  67. ashna
    • 2 years ago
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    yes i do

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

  69. beginnersmind
    • 2 years ago
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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.

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

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

  72. waterineyes
    • 2 years ago
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    This is divisible by 3 now...

  73. beginnersmind
    • 2 years ago
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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.

  74. ashna
    • 2 years ago
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    right YAY :D ty @beginnersmind and @waterineyes

  75. beginnersmind
    • 2 years ago
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    Not finished yet. But almost :)

  76. ashna
    • 2 years ago
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    i can do the rest ty :)

  77. beginnersmind
    • 2 years ago
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    Ok :)

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