that $k\le m$, we have $$\label{eq-2-1}
\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+1\choose 4t-2j+1}=2^{4m-2}+(-1)^m2^{2m-1}.$$

Note that $\{j\mid j\in Z, 0\le j\le 2k\}=\{2i\mid i\in Z, 0\le i\le k\}\cup \{2i+1\mid i\in Z, 0\le i\le k-1\}$. We have $$\begin{array}{lll}
&&\displaystyle\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+1\choose 4t-2j+1}\\
&=&\displaystyle\sum_{i=0}^{k}{4k\choose 4i}\sum_{t=0}^m{4(m-k)+1\choose 4(t-i)+1}+\sum_{i=0}^{k}{4k\choose 4i+2}\sum_{t=0}^m{4(m-k)+1\choose 4(t-i)-1}\\
&=&\displaystyle\sum_{i=0}^{k}{4k\choose 4i}\sum_{t'=-i}^{m-i}{4(m-k)+1\choose 4t'+1}+\sum_{i=0}^{k}{4k\choose 4i+2}\sum_{t'=-i}^{m-i}{4(m-k)+1\choose 4t'-1}\\
&=&\displaystyle\sum_{i=0}^{k}{4k\choose 4i}\sum_{t'=0}^{m-k}{4(m-k)+1\choose 4t'+1}+\sum_{i=0}^{k}{4k\choose 4i+2}\sum_{t'=0}^{m-k}{4(m-k)+1\choose 4t'-1},\\
\end{array}$$ the third equality holds because ${4(m-k)+1\choose 4t'+1}=0$ and ${4(m-k)+1\choose 4t'-1}=0$ if $t'<0$ or $t'>m-k$. From Lemmas \[lem-z-6\] and \[lem-z-7\], we have $$\begin{array}{lll}
&&\displaystyle\sum_{i=0}^{k}{4k\choose 4i}\sum_{t'=0}^{m-k}{4(m-k)+1\choose 4t'+1}+\sum_{i=0}^{k}{4k\choose 4i+2}\sum_{t'=0}^{m-k}{4(m-k)+1\choose 4t'-1}\\
&=&\displaystyle(2^{4k-2}+(-1)^k2^{2k-1})(2^{4(m-k)-1}+(-1)^{m-k}2^{2(m-k)-1})\\&&+(2^{4k-2}-(-1)^k2^{2k-1})(2^{4(m-k)-1}-(-1)^{m-k}2^{2(m-k)-1})\\
&=&\displaystyle2^{4m-2}+(-1)^m2^{2m-1}.
\end{array}$$

It completes the proof.

By the similar methods, we obtain the following combination identities.

\[thm-z-3\] For two positive integers $k$ and $m$ such that $k\le m$, we have\
(i) $\displaystyle 2\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j}{4m-4k-1\choose 4t-2j}=2^{4m-2}+(-1)^m2^{2m-1}$;\
(ii) $\displaystyle \sum_{t=0}^m\sum_{j=0}^{2k}{4k+2\choose 2j+1}{4m-4k-1\choose 4t-2j}=2^{4m-2}+(-1)^m2^{2m-1}$;\
(iii) $\displaystyle 2\sum_{t=0}^m\sum_{j=0}^{2k+1}{4k+3\choose 2j+1}{4m-4k-3\choose 4t-2j-1}=2^{4m-2}+(-1)^m2^{2m-1}$;\
(iv) $\displaystyle \sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+3\choose 4t-2j+1}=2^{4m}+(-1)^m2^{2m}$;\
(v) $\displaystyle 2\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j}{4m-4k+1\choose 4t-2j}=2^{4m}+(-1)^m2^{2m}$;\
(vi) $\displaystyle \sum_{t=0}^m\sum_{j=0}^{2k}{4k+2\choose 2j+1}{4m-4k+1\choose 4t-2j}=2^{4m}+(-1)^m2^{2m}$;\
(vii) $\displaystyle 2\sum_{t=0}^m\sum_{j=0}^{2k+1}{4k+3\choose 2j+1}{4m-4k-1\choose 4t-2j-1}=2^{4m}+(-1)^m2^{2m}$;\
(viii) $\displaystyle \sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+3\choose 4t-2j}=2^{4m}-(-1)^m2^{2m}$;\
(ix) $\displaystyle 2\sum_{t=0}^m\sum_{j=0}^{2k+1}{4k+3\choose 2j}{4m-4k-1\choose 4t-2j}=2^{4m}-(-1)^m2^{2m}$;\
(x) $\displaystyle 2\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j+1}{4m-4k+1\choose 4t-2j-1}=2^{4m}-(-1)^m2^{2m}$;\
(xi) $\displaystyle \sum_{t=0}^m\sum_{j=0}^{2k}{4k+2\choose 2j+1}{4m-4k+1\choose 4t-2j+3}=2^{4m}-(-1)^m2^{2m}$.

As similar as the proof of Theorem \[thm-z-2\], we prove these identities by two steps. At first, by using $\{j\mid j\in Z, 0\le j\le 2k\}=\{2i\mid i\in Z, 0\le i\le k\}\cup \{2i+1\mid i\in Z, 0\le i\le k-1\}$, the 