$$\begin{array}{ll}
8 \cdot 3 & \text { Let the sequence } x_{1}, x_{2}, x_{3}, \cdots \text { satisfy } \\
& 3 x_{n}-x_{n-1}=n, n=2,3, \cdots \\
\text { and } & \left|x_{1}\right|<1971 .
\end{array}$$

Find \( x_{1971} \), accurate to 0.000001.
[Solution] Clearly, the sequence $y_{n}=\frac{1}{2} n-\frac{1}{4}, n=1,2,3, \cdots$ satisfies
$$3 y_{n}-y_{n-1}=n, n=2,3, \cdots$$

Therefore, $\left|y_{n}-x_{n}\right|=\frac{1}{3}\left|y_{n-1}-x_{n-1}\right|$,
from which we can deduce that $\left|y_{1971}-x_{1971}\right|=\left(\frac{1}{3}\right)^{1970}\left|y_{1}-x_{1}\right|<\left(\frac{1}{3}\right)^{1970} \cdot 1972$. Thus, $y_{1971}=985.250000$ is the approximation of $x_{1971}$ accurate to 0.000001.