2. The sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=0, a_{1}=1, a_{n+2}=a_{n+1}+a_{n}(n \in \mathbf{N})$, and the sequence $\left\{b_{n}\right\}$ satisfies $b_{1} \in \mathbf{R}, b_{n}=$ $\frac{a_{n-2} b_{1}-a_{n-1}}{a_{n}-a_{n-1} b_{1}}(n=2,3,4, \cdots)$. If $b_{2009}=\frac{1}{b_{1}}+1$, then $b_{1}=$ $\qquad$ .
2. $\frac{1 \pm \sqrt{5}}{2}$ Detailed Explanation: Let $b_{n}=\frac{1}{b_{1}}+1(n=2009)$, then $\frac{a_{n-2} b_{1}-a_{n-1}}{a_{n}-a_{n-1} b_{1}}=\frac{1}{b_{1}}+1$,

we get $\frac{1}{b_{1}}=\frac{a_{n} b_{1}-a_{n+1}}{a_{n}-a_{n-1} b_{1}}$, then $a_{n} b_{1}^{2}-a_{n} b_{1}-a_{n}=0$, so $b_{1}^{2}-b_{1}-1=0$, then $b_{1}=\frac{1 \pm \sqrt{5}}{2}$.