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### Proof of a combinatorial identity: $\sum_{i=0}^n {2i \choose i}{2(n-i)\choose n-i} = 4^n$ [duplicate]

Possible Duplicate: Identity involving binomial coefficients This was part of a homework assignment that I had, and I couldn't figure it out. Now it is bugging me. Can anyone help me? Although ...
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### Proving $\sum_{k=0}^n{2k\choose k}{2n-2k\choose n-k}=4^n$ [duplicate]

Some background. I was asked to find an arithmetic function $f$ such that $f*f=\mathbf 1$ where $\mathbf 1$ is the constant function 1 and $*$ denotes Dirichlet convolution. I was able to prove that ...
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### Finding a closed form expression for this sum [duplicate]

For non-negative $n$, find $$\sum_{k=0}^n \binom{2k}{k}\binom{2n-2k}{n-k}.$$ I can't figure this out. Any ideas?
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### Proving an Identity involving $4^N$ [duplicate]

I am trying to prove the following identity: $$\sum_{k=0}^N\left({2 \, N - 2 \, k \choose N - k}{2 \, k \choose k}\right)=4^N$$ I have tried writing $4^N=2^{2N}=(1+1)^{2N}=(1+1)^N(1+1)^N$, and ...
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### How prove this $\sum_{k+j=n,0\le k,j\le n}\binom{2k}{k}\binom{2j}{j}=4^n$ [duplicate]

Show that $$\sum_{k\ +\ j\ =\ n\atop{\vphantom{\LARGE A}0\ \le\ k,\phantom{A} j\ \le\ n}}{2k \choose k}{2j \choose j} = 4^{n}$$ I think use integral solve it. But I don't it,and this problem is ...
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### Binomial coefficients identity [duplicate]

Prove algebraically or otherwise: $$\sum \limits_{r=0}^n {2r \choose r} {2n-2r \choose n-r} = 4^n$$ where ${n \choose r}$ denotes the usual binomial coefficient. I think there is a combinatorial ...
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### Combinatorial proof of an binomial coefficient identity [duplicate]

The identity $$\sum_k \binom{2 k}{k} \binom{2n - 2k}{n - k} = 4^n$$ is found on page 187 of "Concrete Mathematics" by Knuth and Graham. The book does not prove it combinatorially. Is there a proof ...