For any $n$ prove the following identity $$ \sum_{i+j+k=n} j \binom{n}{i,j,k} a_{2i+1,2j-1,2k}=\sum_{i+j+k=n} i \binom{n}{i,j,k} a_{2i-1,2j+1,2k}, $$ here $$\binom{n}{i,j,k}=\frac{n!}{i! j! k!}, $$ is the multinomial coefficient, $i, j, k \geq 0.$
For small $n=1,2$ I can do it by direct calculation but what about arbitrary $n?$
We have \begin{align} \sum_{i+j+k = n} j\binom{n}{i,j,k}a_{2i+1,2j-1,2k} &= \sum_{\substack{i+j+k = n \\ j > 0}} j\binom{n}{i,j,k}a_{2i+1,2j-1,2k} \\ &= \sum_{\substack{i+j+k = n \\ j > 0}} n\binom{n-1}{i,j-1,k}a_{2i+1,2j-1,2k} \\ &= \sum_{i + \ell + k = n-1} n \binom{n-1}{i,\ell,k} a_{2i+1,2\ell+1,2k} \tag{$\ell = j-1$}\\ &= \sum_{\substack{m + \ell + k = n \\ m > 0}} n\binom{n-1}{m-1,\ell,k} a_{2m-1,2\ell+1,2k} \tag{$m = i+1$}\\ &= \sum_{m + \ell + k = n} m\binom{n}{m,\ell,k} a_{2m-1,2\ell+1,2k} \\ &= \sum_{i + j + k = n} i\binom{n}{i,j,k} a_{2i-1,2j+1,2k}\,. \tag{$i = m, j = \ell$} \end{align}