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Math Help - Binomial coefficients?

  1. #1
    Super Member fardeen_gen's Avatar
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    Binomial coefficients?

    Let k and n be positive integers and put S(k) = 1^k + 2^k + 3^k + ... + n^k.

    Show that (m + 1)C(1)S(1) + (m + 1)C(2)S(2) + ... + (m + 1)C(m)S(m) = (n + 1)^(m + 1) - (n + 1)

    NOTE: (m + 1)C(1) is (m + 1) choose 1
    S(k) - k is subscript
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  2. #2
    Super Member PaulRS's Avatar
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    In this post we we'll take: 0^0=1 to shorten the thing.

    Note that: <br />
\sum\limits_{j = 1}^n {\left[ {\left( {j + 1} \right)^k  - j^k } \right]}  = \left( {n + 1} \right)^k  - 1<br />

    By the Binomial Theorem: <br />
\sum\limits_{j = 1}^n {\left[ {\sum\limits_{s = 0}^k {\binom{k}{s}\cdot j^s }  - j^k } \right]}  =\sum\limits_{j = 1}^n {\left[ {\sum\limits_{s = 0}^{k-1}{\binom{k}{s}\cdot j^s }} \right]}  = \left( {n + 1} \right)^k  - 1<br />

    Now reverse the Order of the sums: \sum\limits_{s = 0}^{k-1} {\binom{k}{s}\cdot\left[ {\sum\limits_{j = 1}^{n}{ j^s }} \right]} = \left( {n + 1} \right)^k - 1<br />

    Note that for s=0: \binom{k}{s}\cdot \sum\limits_{j = 1}^{n}{ j^s }=n thus \sum\limits_{s = 1}^{k-1} {\binom{k}{s}\cdot\left[ {\sum\limits_{j = 0}^{n}{ j^s }} \right]} = \left( {n + 1} \right)^k - (n+1)<br />
    Last edited by PaulRS; April 23rd 2009 at 02:50 PM. Reason: Had changed the inferior limit when exchanging the summation order
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