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Math Help - factor np out of binomial summation

  1. #1
    Member oldguynewstudent's Avatar
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    factor np out of binomial summation

    I totally screwed this problem up earlier. I understand what needs to be done but it's been 35 years since I studied sequences and series. Could someone help by giving me a hint on how to proceed?

    I need to factor out np from the following sigma summation:

    \sum_{x=1}^{n}\left({n\atop x}\right)xp^{x}(1-p)^{n-x}=np(1-p)^{n-1}+\frac{n!}{2!(n-2)!}2p^{2}(1-p)^{n-2}+\ldots+

    Now factoring out the np is not a problem in itself, and I need to change the limits by letting y=x-1 so I can sum over y=0 to y=n-1, but what do I do with the (1-p)^{n-1} in the first term?

    Thanks
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  2. #2
    Senior Member
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    <br />
\sum_{x = 1} ^ n x \binom{n}{x}p^x (1 - p)^{n - x}<br />

    = np \sum_{x = 1} ^ n \frac{(n-1)!}{(x - 1)! (n-1 - (x - 1))!} p^{x - 1} (1 - p)^{(n - 1) - (x - 1)}<br />

    = np \sum_{x = 0} ^ {n - 1} \binom{n - 1}{x}p^x (1 - p)^{(n - 1) - x}<br />

    I'll leave some of the details between the equalities, as well as the final step, to you.
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