Example 1 (Sichuan Preliminary Contest 2004) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n}=2 a_{n-1}+n-2(n \geqslant 2)$, find the general term $a_{n}$.
Solution: From the given, we have: $a_{n}+n=2\left(a_{n-1}+n-1\right)(n \geqslant 2)$. Let $b_{n}=a_{n}+n$, then $b_{1}=a_{1}+1=2$ and $b_{n}=2 b_{n-1}(n \geqslant 2)$.

Thus, $b_{n}=2 \cdot 2^{n-1}=2^{n}$, i.e., $a_{n}+n=2^{n}$, hence $a_{n}=2^{n}-n(n \geqslant 2)$,

Since $a_{1}=1$ also fits the above formula, we have $a_{n}=2^{n}-n(n \geqslant 1)$.