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Spivak 2-1b, 2-2a (induction)

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    1b) Prove by induction: [itex]1^{3}+...+n^{3}=(1+...+n)^{2}[/itex]
    2a) Find a formula for: [itex]\sum^{n}_{i=1}(2i-1)[/itex]

    2. Relevant equations

    There's a Hint for 2a): 'What to this expression have to do with [itex]1+2+3+...+2n[/itex]?'


    3. The attempt at a solution

    In 2a) I've got near the answer, when comparing with the given one, but I can't understand the last thing he does. The solution in the book is:

    [itex]\sum^{n}_{i=1}(2i-1)=1+2+3+...+2n-2(1+...+n)
    =(2n)(2n+1)/2-n(n+1)[/itex]

    And I couldn't understand how to make the second member become the third one, which goes directly to the answer [itex]n^{2}[/itex]


    Thanks
     
  2. jcsd
  3. Aug 22, 2011 #2

    gb7nash

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    Homework Helper

    It's a well known fact that for positive integer n:

    1+2+3+...+n = n(n+1)/2

    Use this to obtain the answer.
     
  4. Aug 22, 2011 #3
    We can get the odd integers by first starting with all integers and removing those which are even.
     
  5. Aug 22, 2011 #4
    Got it, thanks.
     
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