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Thread: Evaluating a Series of Binomial Co-efficients

  1. #1
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    Evaluating a Series of Binomial Co-efficients

    Hey guys,

    I've been scratching my head over this one:

    Evaluate:
    $\displaystyle \binom{40}{0}+\binom{40}{1}+\binom{40}{2}+\dots+\b inom{40}{40}$

    That is to say,

    $\displaystyle \sum_{n=0}^{40}\binom{40}{n}$

    My first thought was to say

    $\displaystyle \sum_{n=0}^{40}\binom{40}{n} = \sum_{n=0}^{40}\frac{40!}{40(40-n)!} = \sum_{n=0}^{40}\left(\frac{A}{40}+\frac{B}{(40-n)!}\right)$

    Then to find $\displaystyle A$ and $\displaystyle B$ by partial fractions and do a summation table, but couldn't get it out. Any suggestions from anyone?
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  2. #2
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    Quote Originally Posted by Stephen Shaw View Post
    Evaluate:
    $\displaystyle \binom{40}{0}+\binom{40}{1}+\binom{40}{2}+\dots+\b inom{40}{40}$
    That is to say,
    $\displaystyle \sum_{n=0}^{40}\binom{40}{n}$
    Think about it for a moment.
    That is simply the number of subsets of a set of 40 elements: $\displaystyle 2^{40}$.
    Another way. $\displaystyle \left( {x + y} \right)^{40} = \sum\limits_{k = 0}^{40} {{40 \choose k}x^{40 - k} y^k } $
    Let $\displaystyle x=1\;\&\;y=1$ we have $\displaystyle \left( {2} \right)^{40} = \sum\limits_{k = 0}^{40} {{40 \choose k} } $
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