Results 1 to 6 of 6

Math Help - Binomial Expansion Problem - Please help!

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    16

    Binomial Expansion Problem - Please help!

    Hey all this is my first problem i have posted

    here is the question

    In the binomial expansion of (2k+x)^n, where k is a constant and n is a positive integer, the coefficient of x^2 is equal to the coefficient of x^3

    a) Prove that n=6k+2

    Thanks, i really do not know where to start

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by acid_rainbow View Post
    Hey all this is my first problem i have posted

    here is the question

    In the binomial expansion of (2k+x)^n, where k is a constant and n is a positive integer, the coefficient of x^2 is equal to the coefficient of x^3

    a) Prove that n=6k+2

    Thanks, i really do not know where to start

    thanks
    Start by writting down the two coefficients in question:

    Coeff of x^2 : {n \choose 2}(2k)^{n-2}

    Coeff of x^3 : {n \choose 3}(2k)^{n-3}

    So you have:

    {n \choose 2}(2k)^{n-2}= {n \choose 3}(2k)^{n-3}

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    16
    yea i got that far, but how do i solve/ rearrange that to get n=6k+2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2008
    Posts
    16
    how can you simplify something if its NC2 :SSSSS

    so confused
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack View Post
    Start by writting down the two coefficients in question:

    Coeff of x^2 : {n \choose 2}(2k)^{n-2}

    Coeff of x^3 : {n \choose 3}(2k)^{n-3}

    So you have:

    {n \choose 2} (2k)^{n-2} = {n \choose 3}(2k)^{n-3}

    RonL
     <br />
\frac{n!}{2!(n-2)!} \frac{(2k)^{n-2}}{(2k)^{n-3}}=\frac{n!}{3!(n-3)!}<br />

    so:

     <br />
\frac{1}{2!(n-2)!} (2k) = \frac{1}{3!(n-3)!}<br />

     <br />
\frac{1}{(n-2)} (2k) = \frac{1}{3}<br />

    which will rearrange into what you want

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2008
    Posts
    16
    thanks, sorry for double post, im not sure where the 1 comes from but i will just learn it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: November 14th 2010, 12:51 PM
  2. [SOLVED] Binomial expansion problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 22nd 2010, 12:41 PM
  3. Binomial expansion problem
    Posted in the Algebra Forum
    Replies: 0
    Last Post: February 15th 2010, 08:43 AM
  4. Replies: 6
    Last Post: May 1st 2009, 11:37 AM
  5. Binomial Expansion
    Posted in the Algebra Forum
    Replies: 16
    Last Post: July 24th 2008, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum