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Induction involving Sum

  1. May 28, 2006 #1
    Prove that [tex] \sum_{k=1}^{2n} (-1)^{k}(2k+1) [/tex] is proportional to [tex] n [/tex], and find the constant of proportionality. So I have to prove that: [tex] \sum_{k=1}^{2n} (-1)^{k}(2k+1) = np [/tex] where [tex] p [/tex] is the constant of proportionality. So for [tex] n =1 [/tex] we have [tex] \sum_{k=1}^{2} (-1)^{k}(2k+1) = 2 [/tex]. In this case, [tex] p = \frac{1}{2} [/tex]. For [tex] n =s [/tex] we have [tex] \sum_{k=1}^{2s}(-1)^{k}(2k+1) = -3 + 5 + ... + (-1)^{2s}(4s+1) = ps [/tex] which we assume to be true. Then for [tex] n = s+1 [/tex] we have [tex] \sum_{k=1}^{2s+2} (-1)^{k}(2k+1) = p(s+1) [/tex] which we want to prove from the case [tex] n = s [/tex]. So [tex] \sum_{k=1}^{2s+2}(-1)^{k}(2k+1) = (\sum_{k=1}^{2s} (-1)^{k}(2k+1))+(-1)^{2s+1}(4s+3) [/tex].

    Am I doing this correctly? I am stuck at this part.

    Thanks
     
  2. jcsd
  3. May 28, 2006 #2

    Tide

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    You left out a term in your last equation:

    [tex]\sum^{2s+2}a_k = \sum^{2s} a_k + a_{2s+1} + a_{2s+2}[/tex]
     
  4. May 28, 2006 #3
    So that means [tex] \sum_{k=1}^{2n+2} = (\sum_{k=1}^{2s} a_{k}) + a_{2n+1}+a_{2n+2} = ps + (-1)^{2n+1}(4n+3) + (-1)^{2s+2}(4s+5) = ps-4n-3 + 4s + 5 = ps + 2 [/tex]. So [tex] ps+2 = ps + p [/tex] so does that mean [tex] p = 2 [/tex]?

    Thanks
     
  5. May 28, 2006 #4

    VietDao29

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    Yes, p = 2, not 1 / 2. This part is wrong.
    [tex]\sum_{k=1}^{2} (-1)^{k}(2k+1) = 2 = np = 1 \times p = p[/tex]
    So p = 2.
    :)
     
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