7. (1994 Hefei High School Mathematics Competition, Anhui Province) Let $a_{1}, a_{2}, \cdots$ and $b_{1}, b_{2}, \cdots$ be two arithmetic sequences, with their sums of the first $n$ terms being $A_{n}$ and $B_{n}$, respectively. It is known that for all $n \in \mathbf{N}$,
$$
\frac{A_{n}}{B_{n}}=\frac{2 n-1}{3 n+1} \text {. }
$$

Try to write the expression for $\frac{a_{n}}{b_{n}}$ for all $n \in \mathbf{N}$.
7. 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
$$
\begin{aligned}
A_{n} & =n a_{1}+\frac{n(n-1)}{2} d_{1}, B_{n}=n b_{1}+\frac{n(n-1)}{2} d_{2}, \\
\frac{A_{n}}{B_{n}} & =\frac{2 a_{1}+(n-1) d_{1}}{2 b_{1}+(n-1) d_{2}} .
\end{aligned}
$$

From the given conditions, for all $n \in \mathbf{N}$, 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
$$
\frac{a_{1}}{b_{1}}=\frac{1}{4}, b_{1}=4 a_{1} \text {. }
$$

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
$$
\frac{2 a_{1}+d_{1}}{8 a_{1}+\frac{3}{2} d_{1}}=\frac{3}{7} \text {, simplifying to } d_{1}=4 a_{1} \text {. }
$$

Thus, $d_{2}=\frac{3}{2} d_{1}=6 a_{1}$,
so $\frac{a_{n}}{b_{n}}=\frac{a_{1}+(n-1) d_{1}}{b_{1}+(n-1) d_{2}}=\frac{a_{1}+(n-1) \cdot 4 a_{1}}{4 a_{1}+(n-1) \cdot 6 a_{1}}$
$$
=\frac{1+4(n-1)}{4+6(n-1)}=\frac{4 n-3}{6 n-2} .
$$