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Can anyone solve this mathematical induction problem?

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that for all n>1,
    P(n) :[itex]1 + 1/2 + 1/3 +....+1/n = k/m [/itex]

    where k is an odd number an m is an even number.


    2. Relevant equations



    3. The attempt at a solution
    1)Base Case n =2
    P(2) = k/m
    [itex] 3/2 = k/m [/itex] which is true.

    2) Inductive Step
    Assume P(n) is true for some arbitrary n.

    3) Prove P(n+1)
    [itex] P(n+1) = k/m [/itex]
    [itex]P(n) +1/(n+1) = k/m [/itex]
    We know/assume that P(n) has an odd numerator and an even denominator. So,
    [itex] k/m + 1/(n+1) = k/m [/itex]
    [itex] (k(n+1) +m)/ (m+mn) [/itex]

    So i divided the problem into two cases:

    Case 1: n+1 is odd
    if n+1 is odd then k(n+1) + m is odd and m +mn is even which is true.

    Case 2: n+1 is even
    here's where the problem is
     
  2. jcsd
  3. Oct 14, 2012 #2

    rude man

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    Let me rewrite your denominator mn + m = m(n+1).

    Multiply m(n+1) by 2p.

    Show that there exists a positive integer value for p such that the numerator must wind up odd whereas the denominator will always be even.
     
  4. Oct 15, 2012 #3

    CEL

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    k is odd. So, k(n+1) is odd, no matter if (n+1) is odd or even. Since m is even, k(n+1) + m is odd.
     
  5. Oct 15, 2012 #4

    rude man

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    Huh?

    If k is odd and (n+1) is even, k(n+1) is even!

    Odd x even = even always!
     
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