Results 1 to 5 of 5

Math Help - Binomial coefficients 3

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    15

    Binomial coefficients 3

    This problem is very difficult...can someone help me with this please? ...this is all about binomials



    1. Show that


    <br />
\sum_{k=1}^{n}k\binom{n}{2k+1}=(n-2){2}^{n-3}<br />
    Last edited by manusform; September 17th 2009 at 12:26 AM. Reason: equation is not appropriate
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2009
    Posts
    125
    Hello,
     (1-x)^n=\sum_{k=0}^n (-1)^k \binom{n}{k}x^k=-\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1} + \sum_{k=0}^n  \binom{n}{2k}x^{2k}

     (1+x)^n=\sum_{k=0}^n  \binom{n}{k}x^k=\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1  } + \sum_{k=0}^n  \binom{n}{2k}x^{2k}

    Take derivative with respect to x of both equalities

    -n(1-x)^{n-1}=-\sum_{k=0}^n (2k+1) \binom{n}{2k+1}x^{2k} + \sum_{k=0}^n 2k \binom{n}{2k}x^{2k-1}

    n(1+x)^{n-1}=\sum_{k=0}^n (2k+1) \binom{n}{2k+1}x^{2k} + \sum_{k=0}^n 2k \binom{n}{2k}x^{2k-1}

    Set x=1 in all four equalities and you get

    \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n  \binom{n}{2k} =2^{n-1}

    \sum_{k=0}^n (2k+1) \binom{n}{2k+1} = \sum_{k=0}^n 2k \binom{n}{2k} = n2^{n-2} for n>1

    Now because

    \sum_{k=0}^n (2k+1) \binom{n}{2k+1} = 2\sum_{k=0}^n k \binom{n}{2k+1}+\sum_{k=0}^n  \binom{n}{2k+1}

    you get

    \sum_{k=0}^n k \binom{n}{2k+1} = \frac{1}{2}(n2^{n-2} - 2^{n-1}) =\frac{1}{2}(n2^{n-2} - 2\cdot2^{n-2})=(n-2)2^{n-3} for  n>1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2009
    Posts
    36
    I think you made a mistake.

    \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n  \binom{n}{2k} \neq2^{n-1}

    The right form is:

    \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2k  +1} = \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}  \binom{n}{2k} =2^{n-1}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2009
    Posts
    36
    Sorry, both two form are right.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2009
    Posts
    125
    Hi streethot,

    I don't think it is a mistake, since the definition of binomial coefficients says
    \binom{n}{k}=0 for k>n. This means your form and my form are equivalent.

    I gave a derivation of \sum_{k=0}^n\binom{n}{2k+1} = \sum_{k=0}^n \binom{n}{2k} = 2^{n-1} above.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Binomial Coefficients
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: February 1st 2010, 07:24 PM
  2. Binomial coefficients
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 25th 2009, 06:18 PM
  3. Binomial Coefficients
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 16th 2008, 08:04 AM
  4. Binomial coefficients
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: April 5th 2008, 12:21 PM
  5. Binomial Coefficients
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: July 23rd 2007, 11:47 AM

Search Tags


/mathhelpforum @mathhelpforum