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Math Help - Solving sum

  1. #1
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    Solving sum

    How to solve this or simplify? \sum_{k=0}^{n}\binom{n}{k}k^2p^k(1-p)^(n-k)
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  2. #2
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    Re: Solving sum

    Quote Originally Posted by Garas View Post
    How to solve this or simplify? \sum_{k=0}^{n}\binom{n}{k}k^2p^k(1-p)^{n-k}
    Start with the binomial expansion (p+q)^n = \sum_{k=0}^{n}\binom{n}{k}p^kq^{n-k}. Differentiate partially with respect to p (keeping q constant), then multiply both sides by p. That will give np(p+q)^{n-1} = \sum_{k=0}^{n}\binom nk kp^kq^{n-k}. Now do the same thing again: np(p+q)^{n-1} + n(n-1)p^2(p+q)^{n-2} = \sum_{k=0}^{n}\binom nk k^2p^kq^{n-k}. Finally, put q=1-p to get \sum_{k=0}^{n}\binom{n}{k}k^2p^k(1-p)^{n-k} = np+n(n-1)p^2.
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