Seeking a combinatorial proof $\sum _{k=0}^{2n} (-1)^k \binom{4n-2k}{2n-k}\binom{2k}{k}=\binom{2n}{n}\times 2^{2n}$

Question

I would appreciate if somebody could help me with the following problem

Q: Seeking a combinatorial proof $(\binom{n}{k}=\frac{n!}{k! (n-k)!} )$

$$\sum _{k=0}^{2n} (-1)^k \binom{4n-2k}{2n-k}\binom{2k}{k}=\binom{2n}{n}\times 2^{2n}$$

See http://math.stackexchange.com/questions/80649/combinatorial-proof-that-sum-limits-k-0n-binom2kk-binom2n-2kn-k?rq=1 for a similar question. — Nicholas, Oct 25 '15 at 16:29

score 3 · Accepted Answer · answered Oct 25 '15 at 16:31

Since $$ \sum_{k\geq 0}\binom{2k}{k}x^k = \frac{1}{\sqrt{1-4x}}, \tag{1}$$ your sum can be read as a convolution, i.e. as the coefficient of $x^{2n}$ in the product between $\frac{1}{\sqrt{1-4x}}$ and $\frac{1}{\sqrt{1+4x}}$, but, due to $(1)$: $$ [x^{2n}]\frac{1}{\sqrt{1-16 x^2}} = [x^n]\frac{1}{\sqrt{1-16 x}} = \binom{2n}{n}\cdot 4^n,\tag{2}$$ hence: $$ \sum_{k=0}^{2n}\binom{4n-2k}{2n-k}\binom{2k}{k}(-1)^k = \binom{2n}{n}\cdot 2^{2n}. \tag{3}$$

Seeking a combinatorial proof $\sum _{k=0}^{2n} (-1)^k \binom{4n-2k}{2n-k}\binom{2k}{k}=\binom{2n}{n}\times 2^{2n}$

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