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Homework Help: Please check my proof of sum of n choose k = 2^n

  1. Jun 18, 2010 #1
    1. The problem statement, all variables and given/known data
    (here (n,k) reads n choose k)(and again, please excuse that i don't use latex)
    claim: (n,0) + (n,1) + (n,2) + ... (n,n) = 2n


    2. Relevant equations

    binomial theorem

    3. The attempt at a solution
    proof: sum(k=0 to n of (n,k)) = sum(k=0 to n of (n,k))*1k*1n-k.
    by the binomial theorem, (x + y)n = sum(k=0 to n of (n,k))*xkyn-k, so letting x, y = 1, then (1 + 1)n = 2n = sum(k=0 to n of (n,k))*1k*1n-k = sum(k=0 to n of (n,k)).
     
  2. jcsd
  3. Jun 18, 2010 #2

    rock.freak667

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    Looks correct, I see nothing wrong with it. (assuming you weren't given a set method such as mathematical induction to use)
     
  4. Jun 19, 2010 #3
    no. but looking back on it, should i have been a little more rigorous? like being explicitly clear that for the binomial theorem, n has to be a positive integer. it's kind of implied, i guess.
     
  5. Jun 19, 2010 #4

    rock.freak667

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    The actual proof itself looks valid, you can state the conditions before it though.
     
  6. Jun 19, 2010 #5

    Gib Z

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    The theorem in fact remains true for any complex number n if you use generalized binomial coefficients and binomial series. See http://en.wikipedia.org/wiki/Binomial_series
     
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