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Thread: Binomial theorem

  1. #1
    MHF Contributor alexmahone's Avatar
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    Binomial theorem

    If $\displaystyle (1 + x)^n = c_0 + c_1x + c_2x^2 + . . . + c_nx^n$, show that
    $\displaystyle c_0^2 + 2c_1^2 + ... + (n + 1)c_n^2 = \frac {(2n - 1)!(n + 2)}{n!(n - 1)!}$
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  2. #2
    Super Member PaulRS's Avatar
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    That is equivalent to showing that: $\displaystyle
    \sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k ^2 } = n \cdot \binom{2n-1}{n}+ \binom{2n}{n}
    $ (1)

    To prove it: $\displaystyle
    \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} }
    $

    $\displaystyle
    \Rightarrow n \cdot x \cdot \left( {1 + x} \right)^{n - 1} = \sum\limits_{k = 0}^n {c_k \cdot k \cdot x^k }
    $$\displaystyle
    \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 }
    $

    Now consider the coefficient of $\displaystyle x^n$ in the product: $\displaystyle
    \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)
    $

    It is: $\displaystyle
    \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 }
    $ this last equality holds since: $\displaystyle c_{n-k}=c_k$

    NOw, on the other hand: $\displaystyle
    \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
    $$\displaystyle
    = n \cdot x \cdot \left( {1 + x} \right)^{2n - 1} + \left( {1 + x} \right)^{2n}
    $

    So from here we also want the coefficient of $\displaystyle x^n$, 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 $\displaystyle
    \sum\limits_{k = 0}^n {\left( {k + 1} \right) \cdot c_k ^2 } = n \cdot \binom{2n-1}{n}+ \binom{2n}{n}
    $
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  3. #3
    Senior Member pankaj's Avatar
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    Quote Originally Posted by alexmahone View Post
    If $\displaystyle (1 + x)^n = c_0 + c_1x + c_2x^2 + . . . + c_nx^n$, show that
    $\displaystyle c_0^2 + 2c_1^2 + ... + (n + 1)c_n^2 = \frac {(2n - 1)!(n + 2)}{n!(n - 1)!}$
    Let 1) $\displaystyle S=c_0^2 + 2c_1^2 + ... + nc_{n-1}^2+(n + 1)c_n^2$

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

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

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

    Adding 1 and 2 we get

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

    $\displaystyle
    2S=(n+2)C(2n,n)=(n+2) \frac{2n!}{n!n!}
    $
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by pankaj View Post
    $\displaystyle 2S=(n+2)(c_0^2+c_1^2+...+c_{n-1}^2+c_n^2)$

    $\displaystyle
    2S=(n+2)C(2n,n)
    $
    I don't get this step; could you please explain?
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  5. #5
    Senior Member pankaj's Avatar
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    $\displaystyle c_0^2+c_1^2+c_2^2+.......+c_n^2$

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

    This is number of ways to select n objects from 2n objects which is alternatively $\displaystyle C(2n,n).$
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