Results 1 to 5 of 5

Math Help - Binomial theorem

  1. #1
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Binomial theorem

    If (1 + x)^n = c_0 + c_1x + c_2x^2 + . . . + c_nx^n, show that
    c_0^2 + 2c_1^2 + ... + (n + 1)c_n^2 = \frac {(2n - 1)!(n + 2)}{n!(n - 1)!}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    That is equivalent to showing that: <br />
\sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k ^2 }  = n \cdot \binom{2n-1}{n}+ \binom{2n}{n}<br />
(1)

    To prove it: <br />
\left( {1 + x} \right)^n  = \sum\limits_{k = 0}^n {c_k  \cdot x^k } \underbrace  \Rightarrow _{{\rm{differentiate}}}n \cdot \left( {1 + x} \right)^{n - 1}  = \sum\limits_{k = 0}^n {c_k  \cdot k \cdot x^{k - 1} } <br />

    <br />
 \Rightarrow n \cdot x \cdot \left( {1 + x} \right)^{n - 1}  = \sum\limits_{k = 0}^n {c_k  \cdot k \cdot x^k } <br />
<br />
 \Rightarrow n \cdot x \cdot \left( {1 + x} \right)^{n - 1}  + \left( {1 + x} \right)^n  = \sum\limits_{k = 0}^n {c_k  \cdot \left( {k + 1} \right) \cdot x^k } <br />

    Now consider the coefficient of x^n in the product: <br />
\left( {\sum\limits_{k = 0}^n {c_k  \cdot \left( {k + 1} \right) \cdot x^k } } \right) \cdot \left( {\sum\limits_{k = 0}^n {c_k  \cdot x^k } } \right)<br />

    It is: <br />
\sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k  \cdot c_{n - k} }  = \sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k ^2 } <br />
this last equality holds since: c_{n-k}=c_k

    NOw, on the other hand: <br />
\left( {\sum\limits_{k = 0}^n {c_k  \cdot \left( {k + 1} \right) \cdot x^k } } \right) \cdot \left( {\sum\limits_{k = 0}^n {c_k  \cdot x^k } } \right) = \left( {n \cdot x \cdot \left( {1 + x} \right)^{n - 1}  + \left( {1 + x} \right)^n } \right) \cdot \left( {1 + x} \right)^n <br />
<br />
 = n \cdot x \cdot \left( {1 + x} \right)^{2n - 1}  + \left( {1 + x} \right)^{2n} <br />

    So from here we also want the coefficient of x^n, which is exactly the RHS of (1) -by using the binomial theorem-, now, from the uniqueness of the coefficients of the polynomials it follows that <br />
\sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k ^2 }  = n \cdot \binom{2n-1}{n}+ \binom{2n}{n}<br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    Quote Originally Posted by alexmahone View Post
    If (1 + x)^n = c_0 + c_1x + c_2x^2 + . . . + c_nx^n, show that
    c_0^2 + 2c_1^2 + ... + (n + 1)c_n^2 = \frac {(2n - 1)!(n + 2)}{n!(n - 1)!}
    Let 1) S=c_0^2 + 2c_1^2 + ... + nc_{n-1}^2+(n + 1)c_n^2

    S=(n+1)c_n^2+nc_{n-1}^2+...+2c_1^2+c_0^2

    Since. c_r=c_{n-r},we get

    2) S=(n+1)c_0^2+nc_1^2+...+2c_{n-1}^2+c_n^2

    Adding 1 and 2 we get

    2S=(n+2)(c_0^2+c_1^2+...+c_{n-1}^2+c_n^2)

     <br />
2S=(n+2)C(2n,n)=(n+2) \frac{2n!}{n!n!}<br />
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by pankaj View Post
    2S=(n+2)(c_0^2+c_1^2+...+c_{n-1}^2+c_n^2)

     <br />
2S=(n+2)C(2n,n)<br />
    I don't get this step; could you please explain?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    318
    c_0^2+c_1^2+c_2^2+.......+c_n^2

    =c_0c_n+c_1c_{n-1}+c_2c_{n-2}+.......+c_nc_0

    This is number of ways to select n objects from 2n objects which is alternatively C(2n,n).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. binomial theorem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 20th 2010, 02:12 PM
  2. Binomial Theorem or Binomial Coefficient
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 2nd 2009, 01:06 PM
  3. Binomial Theorem?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 24th 2009, 08:52 AM
  4. Binomial Theorem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 5th 2009, 02:31 AM
  5. Binomial theorem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 3rd 2008, 12:44 PM

Search Tags


/mathhelpforum @mathhelpforum