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Binomial Coefficients

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data
    If [tex]\sum^{n}_{r=0} \frac{1}{^{n}C_{r}} = a[/tex], then find the value of [tex]\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}[/tex] in terms of a and n.[/tex]






    3. The attempt at a solution
    I tried to write down the terms of both the series, but to no avail. i cant think of anything.Please shed some light.
     
  2. jcsd
  3. Aug 18, 2008 #2

    tiny-tim

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    Hi ritwik06! :smile:

    Hint: suppose n = 12.

    Then [tex]\sum^{n}_{r=0} \frac{1}{^{n}C_{r}}[/tex]

    = (0!12! + 1!11! + 2!10! + 3!9! + …)/12!

    So what is [tex]\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}[/tex] ? :smile:
     
  4. Aug 18, 2008 #3

    Defennder

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    Hi tim, I'm not seeing how this helps to solve the problem. You have a term dependent r in each summand, so how do we express it in a?
     
  5. Aug 19, 2008 #4
    Thank god! Somebody helped me. But Tim, I wonder what you wish to convey... Please could you be more explicit :smile:
     
  6. Aug 19, 2008 #5

    Dick

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    Consider:
    [tex]
    \sum^{n}_{r=0} \frac{n-r}{^{n}C_{r}}
    [/tex]
    How does that compare with:
    [tex]
    \sum^{n}_{r=0} \frac{r}{^{n}C_{r}}
    [/tex]
    Does that give you any ideas??
     
  7. Aug 22, 2008 #6

    tiny-tim

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    Hi ritwik06!! :smile:

    Have you got this now … you haven't said?

    If you haven't, then follow Dick's hint … it's much better than mine! :redface:

    (same for the other thread)
     
  8. Aug 23, 2008 #7

    Dick

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    That's nice of you to say, tiny-tim. Thanks. :) Now you've got me curious. ritwik06, did you get it? It's surprising easy if you think about it right, and pretty nonobvious if you don't. It took me a while.
     
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