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Math Help - Prove using Binomial Theorem?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove using Binomial Theorem?

    If ∑(r = 0 to n) (nCr)^-1 = s,
    then prove that ∑(r = 0 to n) (2r + 1)/(nCr) = (n+1)s

    nCr is n choose r.
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    If ∑(r = 0 to n) (nCr)^-1 = s,
    then prove that ∑(r = 0 to n) (2r + 1)/(nCr) = (n+1)s

    nCr is n choose r.
    Let \sum_{r=0}^n \frac{1}{{n\choose r}} = s and A = \sum_{r=0}^n \frac{r}{{n\choose r}}.

    Notice that, \sum_{r=0}^n \frac{r}{{n\choose r}} = \sum_{r=0}^n \frac{r}{{n\choose {n-r}}} now let t=n-r and so \sum_{r=0}^n \frac{r}{{n\choose r}} = \sum_{r=0}^n \frac{n-t}{{n\choose t}}.
    Therefore, 2\sum_{r=0}^{n} \frac{r}{{n\choose r}} = \sum_{t=0}^n \frac{n}{{n\choose r}} = ns \implies A = \tfrac{1}{2}ns

    Thus, \sum_{r=0}^n \frac{2r+1}{{n\choose r}} = 2\sum_{r=0}^n \frac{r}{{n\choose r}} + \sum_{r=0}^n \frac{1}{{n\choose r}} = 2\left( \tfrac{1}{2}ns \right) + s = ns + s = s(n+1).
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