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The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found at Bijections for the Identity $4^n = \sum_{k = 0}^n \binom{2k}k\binom{2(n - k)}{n - k}$. $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n. \label{1}\tag1$$ But, I am not sure about the following analogous equation $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\binom{2n}n\frac{2n+1}{2k+1}=4^{2n}. \label{2}\tag2$$ So, I like to ask:

QUESTION. Can you provide alternative proofs (algebraic, combinatorial, etc) to the identity \eqref{2}?

REMARK 1. As an aside, one may consult this discussion by Fedor Petrov on a $q$-analogue of \eqref{1} in an answer to Looking for a $q$-analogue of a binomial identity.

REMARK 2. Here is an equivalent fomulation of \eqref{2}: $$\sum_{k=0}^n\frac{\binom{n}k^2\binom{2n}n^2}{\binom{2n}{2k}}\frac{2n+1}{2k+1}=4^{2n}.$$

The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found at Bijections for the Identity $4^n = \sum_{k = 0}^n \binom{2k}k\binom{2(n - k)}{n - k}$. $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n. \label{1}\tag1$$ But, I am not sure about the following analogous equation $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\binom{2n}n\frac{2n+1}{2k+1}=4^{2n}. \label{2}\tag2$$ So, I like to ask:

QUESTION. Can you provide a variety of proofs (algebraic, combinatorial, etc) to the identity \eqref{2}?

REMARK 1. As an aside, one may consult this discussion by Fedor Petrov on a $q$-analogue of \eqref{1} in an answer to Looking for a $q$-analogue of a binomial identity.

REMARK 2. Here is an equivalent fomulation of \eqref{2}: $$\sum_{k=0}^n\frac{\binom{n}k^2\binom{2n}n^2}{\binom{2n}{2k}}\frac{2n+1}{2k+1}=4^{2n}.$$

The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found at Bijections for the Identity $4^n = \sum_{k = 0}^n \binom{2k}k\binom{2(n - k)}{n - k}$. $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n. \label{1}\tag1$$ But, I am not sure about the following analogous equation $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\binom{2n}n\frac{2n+1}{2k+1}=4^{2n}. \label{2}\tag2$$ So, I like to ask:

QUESTION. Can you provide alternative proofs (algebraic, combinatorial, etc) to the identity \eqref{2}?

REMARK 1. As an aside, one may consult this discussion by Fedor Petrov on a $q$-analogue of \label{1} in an answer to Looking for a $q$-analogue of a binomial identity.

REMARK 2. Here is an equivalent fomulation of \eqref{2}: $$\sum_{k=0}^n\frac{\binom{n}k^2\binom{2n}n^2}{\binom{2n}{2k}}\frac{2n+1}{2k+1}=4^{2n}.$$

The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found at Bijections for the Identity $4^n = \sum_{k = 0}^n \binom{2k}k\binom{2(n - k)}{n - k}$. $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n. \label{1}\tag1$$ But, I am not sure about the following analogous equation $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\binom{2n}n\frac{2n+1}{2k+1}=4^{2n}. \label{2}\tag2$$ So, I like to ask:

QUESTION. Can you provide alternative proofs (algebraic, combinatorial, etc) to the identity \eqref{2}?

REMARK 1. As an aside, one may consult this discussion by Fedor Petrov on a $q$-analogue of \eqref{1} in an answer to Looking for a $q$-analogue of a binomial identity.

REMARK 2. Here is an equivalent fomulation of \eqref{2}: $$\sum_{k=0}^n\frac{\binom{n}k^2\binom{2n}n^2}{\binom{2n}{2k}}\frac{2n+1}{2k+1}=4^{2n}.$$

The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found here. $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n. \tag1$$ But, I am not sure about the following analogous equation $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\binom{2n}n\frac{2n+1}{2k+1}=4^{2n}. \tag2$$ So, I like to ask:

QUESTION. Can you provide alternative proofs (algebraic, combinatorial, etc) to the identity (2)?

REMARK 1. As an aside, one may consult this discussion by Fedor Petrov on a $q$-analogue of (1).

REMARK 2. Here is an equivalent fomulation of (2): $$\sum_{k=0}^n\frac{\binom{n}k^2\binom{2n}n^2}{\binom{2n}{2k}}\frac{2n+1}{2k+1}=4^{2n}.$$

The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found at Bijections for the Identity $4^n = \sum_{k = 0}^n \binom{2k}k\binom{2(n - k)}{n - k}$. $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n. \label{1}\tag1$$ But, I am not sure about the following analogous equation $$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\binom{2n}n\frac{2n+1}{2k+1}=4^{2n}. \label{2}\tag2$$ So, I like to ask:

QUESTION. Can you provide alternative proofs (algebraic, combinatorial, etc) to the identity \eqref{2}?

REMARK 1. As an aside, one may consult this discussion by Fedor Petrov on a $q$-analogue of \label{1} in an answer to Looking for a $q$-analogue of a binomial identity.

REMARK 2. Here is an equivalent fomulation of \eqref{2}: $$\sum_{k=0}^n\frac{\binom{n}k^2\binom{2n}n^2}{\binom{2n}{2k}}\frac{2n+1}{2k+1}=4^{2n}.$$