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Summation proof

  1. Jan 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Does anyone know a clever way to prove that

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

    where B(n,i) is n take i?

    I can do it, but I had to divide into the cases of n = odd and n = even and it took about 1 page front and back. I'm sure there is a trick.

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jan 13, 2008
  2. jcsd
  3. Jan 13, 2008 #2

    morphism

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    Take the second derivative of (1+x)^n and its binomial expansion, then mess with the index of the summation and set x=1. (I'm assuming your n+1 is actually n-1.)
     
  4. Jan 13, 2008 #3

    HallsofIvy

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    ??? It's obviously not true if you replace (n+1) by (n-1). Take n= 1. Then the lefthand side is 1. [iotex]n(n+1)2^{n-2}[/itex] becomes, for n= 1, [itex]1(2)2^{-1}= 1[/itex]. If you replace (n+1) by (n-1), it becomes [tex]2(0)2^{-1}= 0[/itex].
     
  5. Jan 13, 2008 #4

    morphism

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    You're right of course. I was a bit careless with my algebra. The identity I had in mind was:
    [tex]\sum_{i=0}^n i(i-1) {n \choose i} = n(n-1)2^{n-2}[/tex]

    But no worries - a similar trick can still be applied: Differentiate, multiply by x, and differentiate again!
     
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