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Induction question

  1. Sep 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the statement holds for all positive integers n

    [tex] 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{n-1} = 2^n - 1 [/tex]



    2. Relevant equations



    3. The attempt at a solution

    Assume:
    [tex] 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1} = 2^k - 1 [/tex] is true.

    Then:
    [tex] 2^{k+1}-1 = 2^k(2) - 1 = 2(2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1}) [/tex]

    Not sure what to do next :/

    Show that:
    [tex]2^k(2) - 1 = 2(2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1}) [/tex] is true
     
  2. jcsd
  3. Sep 25, 2009 #2

    tiny-tim

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    Homework Helper

    Hi zeion! :smile:

    You have the right basic idea, but you're making it too complicated. :redface:

    Do the obvious :wink:

    just multiply both sides of the original equation by 2 (and then fiddle about with it). :smile:
     
  4. Sep 25, 2009 #3
    Haven't I already proved it?

    Since:
    [tex]
    2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1} = 2^k - 1
    [/tex]

    Then:
    [tex]
    (2)(2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1}) = (2)2^k - 1 = 2^{k+1}-1
    [/tex]
     
  5. Sep 26, 2009 #4

    tiny-tim

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    No, it isn't … be careful! :wink:

    [tex](2)(2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{k-1}) = (2)(2^k - 1) = \cdots\ ?[/tex] :smile:
     
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