3. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=6$, and for any positive integer $n, a_{n+1}+n+1=2\left(a_{n}+1\right)$, then the unit digit of $a_{1}+a_{2}+\cdots+a_{2022}$ is $\qquad$.
$$
a_{n+1}+k(n+1)+b=2\left(a_{n}+k n+b\right) \Rightarrow a_{n+1}=2 a_{n}+k n+b-k \Rightarrow k=-1, b=0 \text {, }
$$

Then $a_{n+1}-(n+1)=2\left(a_{n}-n\right) \Rightarrow a_{n}-n=\left(a_{1}-1\right) \cdot 2^{n-1} \Rightarrow a_{n}=5 \cdot 2^{n-1}+n$. 
Thus, $a_{1}+a_{2}+\cdots+a_{2022}=5\left(1+2+\cdots+2^{2021}\right)+1+2+\cdots+2022$ $=5\left(2^{2022}-1\right)+2023 \times 1011=5 \cdot 2^{2022}+2023 \times 1011-5$,
Therefore, the units digit of $a_{1}+a_{2}+\cdots+a_{2022}$ is 8.