Results 1 to 2 of 2

Math Help - Evaluating a Series of Binomial Co-efficients

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    9

    Evaluating a Series of Binomial Co-efficients

    Hey guys,

    I've been scratching my head over this one:

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

    That is to say,

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

    My first thought was to say

    \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 A and B by partial fractions and do a summation table, but couldn't get it out. Any suggestions from anyone?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Quote Originally Posted by Stephen Shaw View Post
    Evaluate:
    \binom{40}{0}+\binom{40}{1}+\binom{40}{2}+\dots+\b  inom{40}{40}
    That is to say,
    \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: 2^{40}.
    Another way. \left( {x + y} \right)^{40}  = \sum\limits_{k = 0}^{40} {{40 \choose k}x^{40 - k} y^k }
    Let x=1\;\&\;y=1 we have \left( {2} \right)^{40}  = \sum\limits_{k = 0}^{40} {{40 \choose k} }
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stuck on evaluating a series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 14th 2010, 09:24 PM
  2. Replies: 6
    Last Post: April 16th 2010, 08:32 AM
  3. Evaluating series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 10th 2010, 06:06 PM
  4. Replies: 3
    Last Post: March 31st 2010, 07:57 PM
  5. Evaluating Series.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 14th 2008, 05:30 PM

Search Tags


/mathhelpforum @mathhelpforum