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Math Help - Double summation - Binomial Theorem?

  1. #1
    Super Member fardeen_gen's Avatar
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    Double summation - Binomial Theorem?

    Find the value of:
    (i. nCj + j. nCi)
    1≤ i < j ≤ n
    NOTE: i. nCj is i multiplied by n choose j
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Find the value of:
    (i. nCj + j. nCi)
    1≤ i < j ≤ n
    NOTE: i. nCj is i multiplied by n choose j
    In that sum, each integer from 1 to n is multiplied by each binomial coefficient \textstyle n\choose k (k going from 1 to n), except for the "diagonal" products k{n\choose k}. So the sum is equal to \sum_{i=1}^ni\sum_{j=1}^n{n\choose j} - \sum_{k=1}^nk{n\choose k}. But \sum_{i=1}^ni = \tfrac12n(n+1), \sum_{j=1}^n{n\choose j} = 2^n-1 and \sum_{k=1}^nk{n\choose k} = 2^{n-1}n. Therefore \mathop{\sum\sum}_{1\leqslant i<j\leqslant n}\left(i{n\choose j} + j{n\choose i}\right) = \tfrac12n(n+1)(2^n-1) - 2^{n-1}n = 2^{n-1}n^2 - \tfrac12n(n+1).
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