8. In the sequence $\left\{a_{n}\right\}$, if $a_{n+1}+a_{n}=2 n\left(n \in \mathbf{N}_{+}\right)$, then $a_{1}, a_{3}, a_{5}, a_{7}, \cdots$, $a_{2 n-1}, a_{2 n+1}, \cdots$ form an arithmetic sequence with a common difference of 2. Analogously, in the sequence $\left\{b_{n}\right\}$, if $b_{n} \cdot b_{n+1}=3^{n}\left(n \in \mathbf{N}_{+}\right)$, then the conclusion is $\qquad$.
8. $b_{1}, b_{3}, b_{5}, \cdots, b_{2 n-1}, b_{2 n+1}, \cdots$ form a geometric sequence with a common ratio of 3. Given $b_{n} \cdot b_{n+1}=3^{n}$, then $b_{n-1} \cdot b_{n}=3^{n-1}$, hence $\frac{b_{n+1}}{b_{n-1}}=3$. Therefore, $\frac{b_{2 n+1}}{b_{2(n-1)+1}}=3$.