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Math Help - Binomial coefficients

  1. #1
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    Binomial coefficients

    Oh someone pls help..i have hard time solving this but i can't get it...pls help me

    1.show that for n greater than or equal to 0,


    <br />
\sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}<br /> <br />
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  2. #2
    Junior Member
    Joined
    May 2009
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    In general, the product of two polynomials with degrees m and n, respectively, is given by:

    \left(\sum^{m}_{i=0}a_ix^i\right)\left(\sum^{n}_{j  =0}b_jx^j\right)=\sum_{r=0}^{m+n}\left(\sum_{k=0}^  {r}a_kb_{r-k}\right)x^r

    so, using (x+1)^n(x+1)^m=(x+1)^{m+n} we have:

    (x+1)^{m+n}=\sum_{r=0}^{m+n}{m+n\choose r}x^r

    \sum_{r=0}^{m+n}{m+n\choose r}x^r=(x+1)^m(x+1)^n=\left(\sum^{m}_{i=0}{m \choose i}x^i\right)\left(\sum^{n}_{j=0}{n \choose j}x^j\right)= \sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}{m \choose k}{n \choose r-k}\right)x^r

    if
    \sum_{r=0}^{m+n}{m+n\choose r}x^r=\sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}{m \choose k}{n \choose r-k}\right)x^r

    so:

    {m+n\choose r}=\sum_{k=0}^{r}{m \choose k}{n \choose r-k}

    and

    {n \choose k}={n \choose n-k}

    {m+n\choose r}=\sum_{k=0}^{r}{m \choose k}{n \choose k}

    When m=n and r=n we have:

    {2n \choose n}=\sum_{k=0}^{n}{n \choose k}^2
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