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Math Help - Binomial coefficients 4

  1. #1
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    Binomial coefficients 4

    *Show that

    <br />
\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^2=(-1)^n\binom{2n}{n}<br />
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  2. #2
    Member
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    Aug 2009
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    Hi,

    \sum_{k=0}^r (-1)^k\binom{r}{k}x^{2k} = (1-x^2)^r = (1-x)^r(1+x)^r = \sum_{j=0}^r (-1)^j\binom{r}{j}x^{j}\cdot \sum_{k=0}^r \binom{r}{k}x^{k}=
    =\sum_{k=0}^{2r}\left( \sum_{j=0}^k (-1)^j \binom{r}{j}\binom{r}{k-j}  \right)x^k

    Looking at coefficients at x^{2n} we get

    \sum_{j=0}^{2n} (-1)^j \binom{r}{j}\binom{r}{2n-j}=  (-1)^n\binom{r}{n}

    Special case of this is when you take r=2n:

    \sum_{j=0}^{2n} (-1)^j \binom{2n}{j}\binom{2n}{2n-j}=  (-1)^n\binom{2n}{n}

    and using the identity \binom{n}{k}=\binom{n}{n-k} gives you the desired result.
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