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Proof by Induction Problem

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data

    So we have to prove that [itex]\frac{(n+1)(n+2)(n+3)...(2n)}{1*3*5...*(2n-1)}[/itex] = 2n


    2. The attempt at a solution

    I. For n=1, obviously the proposition is true. (2*1/(2-1) = 2^1 = 2)

    II. Let n=k and assume [itex]\frac{(k+1)(k+2)(k+3)...(2k)}{1*3*5...*(2k-1)}[/itex] = 2k.

    Now, for n=k+1 we have: 2k * [itex]\frac{(2k +2)}{2k+1)}[/itex] = 2k+1 → 2(k+1)*[itex]\frac{(k+1)}{2k+1)}[/itex] = 2k+1. Which is not true.

    So, I cannot figure out what am I doing wrong.
    Thanks in advance...
     
  2. jcsd
  3. Oct 15, 2011 #2
    You forgot that in the nominator, you're starting at n+1. So, by doing the transition k->k+1, you need to devide by k+1 also to take that into account. Also, you need to multiply by (2k+1)*(2k+2), since in the nominator you have the product over all numbers between k+1 and 2k.
    So, what you get is: (2k+1)*(2k+2)/(k+1)/(2k+1)=2
     
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