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Prove the following identity by mathematical induction

  1. Jul 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove the following identity by mathematical induction:
    [tex]\sum_{i=1}^n \frac{1}{(2i - 1)(2i + 1)} = \frac{n}{(2n + 1)}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Let P(n) = [tex]\sum_{i=1}^n \frac{1}{(2(1) - 1)(2(1) + 1)} = \frac{1}{(2(1) + 1)}[/tex]

    P(1) = [tex]\sum_{i=1}^n \frac{1}{(1)(3)} = \frac{1}{3}[/tex] is true

    Assuming P(k) is true, then P(k + 1) is also true.

    [tex]\sum_{i=1}^n \frac{1}{(2(k + 1) - 1)(2(k + 1) + 1)} = \frac{k + 1}{(2(k + 1) + 1)}[/tex]

    Do I just change all the i's and n's to k+1 and expand it until left equation is equal to the right?
     
  2. jcsd
  3. Jul 11, 2008 #2
    Re: Induction

    You assume that this is true:
    [tex]\sum_{i=1}^k \frac{1}{(2i - 1)(2i + 1)} = \frac{k}{(2k + 1)}[/tex]

    And then find a way to change the above in to this:

    [tex]\sum_{i=1}^{k+1} \frac{1}{(2i - 1)(2i + 1)} = \frac{k+1}{(2(k+1) + 1)}[/tex]

    Notice how I made the replacement everywhere.
     
  4. Jul 11, 2008 #3

    Defennder

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    Homework Helper

    Re: Induction

    The expression you have there doesn't make sense. The summation is for an index of i, yet k is used. Replacing i with k would still be incorrect. Instead, find a way to write [tex]\sum^{k+1}_{k=0} \frac{1}{(2k+1)(2k-1)} \ \mbox{in terms of} \ \sum^{k}_{k=0} \frac{1}{(2k+1)(2k-1)} \ \mbox{and the (k+1)th summand}[/tex] and then use the induction assumption.

    PS. Avoid using "i" when you are not working with complex numbers. It gets confusing after a while.
     
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