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Math Help - Binomial Question

  1. #1
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    Binomial Question

    Show that for all positiver integers n,

    x[ (1+x)^n-1 + (1+x)^n-2 +...+ (1+x)^2 + (1+x) + 1 ] = (1+x)^n-1


    ii) Hence show that for 1< and= k < and= n,

    (n-1)C(k-1) + (n-2)C(k-2) + (n-3)C(k-3) + ... + (k-1)C(k-1) = nCk


    is nCr (the combinations thing)

    I can't do the second part. Help pls
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  2. #2
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    Quote Originally Posted by noobonastick View Post
    Show that for all positiver integers n,

    x[ (1+x)^n-1 + (1+x)^n-2 +...+ (1+x)^2 + (1+x) + 1 ] = (1+x)^n-1


    ii) Hence show that for 1< and= k < and= n,

    (n-1)C(k-1) + (n-2)C(k-2) + (n-3)C(k-3) + ... + (k-1)C(k-1) = nCk


    is nCr (the combinations thing)

    I can't do the second part. Help pls
    HI

    For part (1) , Its a geometric progression with first term , a=(1+x)^{n-1} , r=(1+x)^{-1}

    So using the summation formula ,

    \frac{(1+x)^{n-1}[1-(1+x)^{-n}]}{1-(1+x)^{-1}}

    \frac{(1+x)^{n-1}[1-(1+x)^{-n}]}{x(1+x)^{-1}}

     <br />
\frac{(1+x)^n\cdot (1+x)^{-1}[\frac{(1+x)^n-1}{(1+x)^n}]}{x(1+x)^{-1}}<br />

    Do some cancellings , then you are left with

    \frac{(1+x)^n-1}{x}

    Both the x's (there is one more x above which i din include in my working) would cancel off .

    Sorry , just realised you are asking help on the second part .. i am gonna post this anyways .
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  3. #3
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    Quote Originally Posted by noobonastick View Post
    Show that for all positiver integers n,

    x[ (1+x)^n-1 + (1+x)^n-2 +...+ (1+x)^2 + (1+x) + 1 ] = (1+x)^n-1


    ii) Hence show that for 1< and= k < and= n,

    (n-1)C(k-1) + (n-2)C(k-2) + (n-3)C(k-3) + ... + (k-1)C(k-1) = nCk


    is nCr (the combinations thing)

    I can't do the second part. Help pls
    For the second part, find the coefficient of x^k on the LHS and RHS of the identity in the first part.
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