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- #1
hxthanh
New member
- Sep 20, 2013
- 16
Evaluate sum:
$\displaystyle S=\sum_{k=0}^{2n}(-1)^k{2n\choose k}{4n\choose 2k}$
$\displaystyle S=\sum_{k=0}^{2n}(-1)^k{2n\choose k}{4n\choose 2k}$
I believe that the answer must be \(\displaystyle S = \frac{(-1)^n(6n)!(2n)!}{(4n)!(3n)!n!}\), but I have NO idea how one might prove that.Evaluate sum:
$\displaystyle S_n=\sum_{k=0}^{2n}(-1)^k{2n\choose k}{4n\choose 2k}$
Your result is absolute correct!I believe that the answer must be \(\displaystyle S = \frac{(-1)^n(6n)!(2n)!}{(4n)!(3n)!n!}\), but I have NO idea how one might prove that.
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