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Proof by Induction

  1. Jul 15, 2011 #1
    The problem statement, all variables and given/known data
    Prove by mathematical induction:
    [itex]\sum^n_{i=1} 1/((2i-1)(2i+1)) = n/(2n+1)[/itex]

    The attempt at a solution
    [itex]\sum^k+1_{i=1} = 1/((2k-1)(2k+1)) + 1/((2(k+1)-1)(2(k+1)+1))[/itex]

    = [itex]k/(2k+1) + 1/(4(k+1)^2 + 2(k+1) - 2(k+1) -1)[/itex]

    = [itex]k/(2k+1) + 1/(4(k+1)^2 -1)[/itex]

    = [itex]k/(2k+1) + 1/(4(k^2 + 2k + 1))[/itex]

    = [itex]k/(2k+1) + 1/(4k^2 8k + 3) - 1)[/itex]

    Solved (thanks Dick):

    = [itex]k/(2k+1) + 1/((2k+3)(2k+1))[/itex]

    = [itex]k(2k+3)/(2k+3)(2k+1) + 1/((2k+3)(2k+1))[/itex]

    = [itex](k(2k+3) + 1)/(2k+3)(2k+1)[/itex]

    = [itex](2k^2 + 3k +1)/(2k+3)(2k+1)[/itex]

    = [itex]((k+1)(2k+1))/(2k+3)(2k+1)[/itex]

    = [itex](k+1)/(2k+3)[/itex]

    = [itex](k+1)/(2k+2+1)[/itex]

    = [itex](k+1)/(2(k+1)+1)[/itex]
     
    Last edited: Jul 15, 2011
  2. jcsd
  3. Jul 15, 2011 #2

    Dick

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    You want to show your result is equal to (k+1)/(2*(k+1)+1). You are going to have a hard time doing that because there's a mistake. 1/(4*(k+1)^2-1) is not equal to 1/(4k^2+2k).
     
  4. Jul 15, 2011 #3
    Thanks; I've fixed that problem; durh.

    Still have no idea how to get from

    1/(4k^2+8k+3)−1) (or 1/((2k+3)(2k+1)) ) to (k+1)/(2*(k+1)+1)
     
  5. Jul 15, 2011 #4

    Dick

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    You want to go from k/(2k+1)+1/((2k+1)*(2k+3)) to (k+1)/(2*(k+1)+1) or (k+1)/(2k+3). It's just algebra. Put everything over a common denominator. Look for common factors to cancel, etc. You don't need to think hard about it before you try just turning the crank and see what happens.
     
  6. Jul 15, 2011 #5
    Thanks for the help Dick.
     
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