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Sum with factorial

  1. Jul 29, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\sum_{n=0}^{100} 1/n!(100-n)![/itex]

    3. The attempt at a solution
    Other then obvious attempts to make sense of the equation's incremental and decrements divisor, I can't figure out where to start with this question. Some assistance would be greatly appreciated.
     
  2. jcsd
  3. Jul 29, 2011 #2

    Dick

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    It looks like it's closely related to the sum of binomial coefficients. What's the sum of C(100,n) for n from 0 to 100? Is that enough of a hint?
     
  4. Jul 29, 2011 #3
    C(100,n) being = 100!/n!(100-n)! ? hmmm, unfortunately I don't see where that's going =/

    Man this one is throwing me for a loop, only question I haven't managed and due tomorrow.
     
  5. Jul 29, 2011 #4

    Dick

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    Yes, that's C(100,n). There is a simple formula for the sum of the binomial coefficients. It's related to the value of (1+1)^100. Don't know it? Expand (1+1)^100 using the binomial theorem.
     
  6. Jul 29, 2011 #5
    well; thanks. But I'm still just totally lost. (1+1)^100 is a massive equation when expanded.
     
  7. Jul 29, 2011 #6
    Unless... x/y = 0 or 1 in the binomial theorem? That would make it easy.

    Makes it = 1/100! ?
     
  8. Jul 29, 2011 #7

    Dick

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    Oh come on, (1+1)^100=2^100. That's an easy enough number to write down. Now what does that have to do with the sum of the binomial coefficients C(100,n)?? C(100,0)+C(100,1)+...+C(100,100). I'm not asking you to evaluate each one. Just think about it.
     
  9. Jul 29, 2011 #8
    wait;

    2^100/100!
     
  10. Jul 29, 2011 #9
    wait;

    2^100/100!
     
  11. Jul 29, 2011 #10

    Dick

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    You aren't just guessing, I hope.
     
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