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Math Help - binomial Series Sum.

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    binomial Series Sum.

    Prove that \binom{n}{0}.\binom{n}{0}-\binom{n+1}{1}.\binom{n}{1}+\binom{n+2}{2}.\binom{  n}{2}+......................+(-1)^n.\binom{2n}{n}.\binom{n}{n}=(-1)^n
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    Why.....
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    MHF Contributor chisigma's Avatar
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    First we recall the identity...

    \displaystyle \sum_{i=0}^{p} \binom{m}{i}\ \binom {n}{p-i} = \binom {m+n}{p} (1)

    ... and then we consider the binomial expansion...

    \displaystyle (1+2\ x)^{n} = \sum_{k=0}^{n} \binom{n}{k} (2\ x)^{k} (2)

    Because is...

    \displaystyle 2^{k} = \sum_{i=0}^{k} \binom{k}{i} (3)

    ... taking into account (1) we can write...

    \displaystyle (1+2\ x)^{n} = \sum_{k=0}^{n} \binom {n}{k}\ x^{k}\ \sum_{i=0}^{k} \binom {k}{i} = \sum_{k=0}^{n} \binom{n+k}{k}\ \binom{n}{k}\ x^{k} (4)

    Now the only You have to do to obtain Your result is setting in (4) x=-1...

    Kind regards

    \chi \sigma
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