8. 81 Let $a_{1}, a_{2}, \cdots$ and $b_{1}, b_{2}, \cdots$ be two arithmetic sequences, and their sums of the first $n$ terms are $A_{n}$ and $B_{n}$, respectively. It is known that for all $n \in N$,
$$\frac{A_{n}}{B_{n}}=\frac{2 n-1}{3 n+1}$$

Try to write the expression for $\frac{a_{n}}{b_{n}}$ for all $n \in N$.
[Solution] Let the common difference of $a_{1}, a_{2}, \cdots$ be $d_{1}$; the common difference of $b_{1}, b_{2}, \cdots$ be $d_{2}$. Then
$$\begin{array}{l}
A_{n}=n a_{1}+\frac{n(n-1)}{2} d_{1} \\
B_{n}=n b_{1}+\frac{n(n-1)}{2} d_{2} \\
\frac{A_{n}}{B_{n}}=\frac{2 a_{1}+(n-1) d_{1}}{2 b_{1}+(n-1) d_{2}}
\end{array}$$

From the given condition, for all $n \in N$, we have
$$\frac{2 a_{1}+(n-1) d_{1}}{2 b_{1}+(n-1) d_{2}}=\frac{2 n-1}{3 n+1} \text {. }$$

In (1), let $n=1$ to get
$$\frac{a_{1}}{b_{1}}=\frac{1}{4}, \quad b_{1}=4 a_{1}$$

In (1), let $n \rightarrow+\infty$ to get
$$\frac{d_{1}}{d_{2}}=\frac{2}{3}, d_{2}=\frac{3}{2} d_{1}$$

In (1), let $n=2$ to get
$$\frac{2 a_{1}+d_{1}}{2 b_{1}+d_{2}}=\frac{3}{7}$$

Using (2) and (3), the above equation becomes
$$\frac{2 a_{1}+d_{1}}{8 a_{1}+\frac{3}{2} d_{1}}=\frac{3}{7}$$

Simplifying, we get
$$d_{1}=4 a_{1}$$

Thus,
$$d_{2}=\frac{3}{2} d_{1}=6 a_{1}$$

Therefore,
$$\begin{aligned}
\frac{a_{n}}{b_{n}} & =\frac{a_{1}+(n-1) d_{1}}{b_{1}+(n-1) d_{2}} \\
& =\frac{a_{1}+(n-1) \cdot 4 a_{1}}{4 a_{1}+(n-1) \cdot 6 a_{1}} \\
& =\frac{1+4(n-1)}{4+6(n-1)} \\
& =\frac{4 n-3}{6 n-2}
\end{aligned}$$