Let $x=(x_0,...,x_{n-1})$ be a word of length $n$ with $x_i\in\{0,1,2\}$. Moreover, let $j(x)=\sum_{i=0}^{n-1}x_i$. Then, of course, $0\leq j(x)\leq 2n$.
Is there a way to "compute" $$ \frac{1}{4^n}\sum_{m=0}^{2n}\sum_{m_1+m_2+m_3=n\\m_2+2m_3=m}2^{m_1}\binom{n}{m_1,m_2,m_3}\lambda^{2n}? $$
Here $m_1$ is the number of zeros, $m_2$ the number of ones and $m_2$ the number of twos occuring in a word of length $n$.
I think I can put the $\lambda^{2n}$ in front of the sums, i.e. get $$ \frac{\lambda^{2n}}{4^n}\sum_{m=0}^{2n}\sum_{m_1+m_2+m_3=n\\m_2+2m_3=m}2^{m_1}\binom{n}{m_1,m_2,m_3}, $$ so that it remains to compute
$$ \sum_{m=0}^{2n}\sum_{m_1+m_2+m_3=n\\m_2+2m_3=m}2^{m_1}\binom{n}{m_1,m_2,m_3} $$
My first impression was that this sum might be $4^n$, since, to my opinion, we can also express this sum as $$ \sum_{m=0}^{2n}\sum_{m_0+m_1+m_2+m_3=n\\m_2+2m_3=m}\binom{n}{m_0,m_1,m_2,m_3} $$ by introducing another letter, say $0'$ whose number is $m_0$. This should replace the factor $2^{m_1}$ in the original sum; this factor should mean that for each letter 0, we have two options. So why not introducing $0'$ as "another version" of $0$, i.e. going over to alphabet $\{0,0',1,2\}$.
With respect to this new alphabet, the inner sum is the coefficient of $x^m$ in the polynomial $(1+1+x+x^2)^n$, isn't it? Call this coefficient $C(n,m)$. Then the sum should be $$ \sum_{m=0}^{2n}C(n,m)=4^n. $$
So my claim is that $$ \frac{1}{4^n}\sum_{m=0}^{2n}\sum_{m_1+m_2+m_3=n\\m_2+2m_3=m}2^{m_1}\binom{n}{m_1,m_2,m_3}\lambda^{2n}=\frac{1}{4^n}\sum_{m=0}^{2n}C(n,m)\lambda^{2n}=\lambda^{2n}. $$
It would be really nice to get some feedback.