6. Define the sequence $\left\{a_{n}\right\}: a_{1}=1, a_{2}=2, a_{n+2}=a_{n}+a_{n+1}, n \in \mathbf{N}_{+}$, then $\left[\frac{a_{2}}{a_{1}}\right] \cdot\left\{\frac{a_{3}}{a_{2}}\right\} \cdot\left\{\frac{a_{4}}{a_{3}}\right\} \cdot \cdots$
- $\left\{\frac{a_{99}}{a_{98}}\right\} \cdot\left[\frac{a_{98}}{a_{2}}\right]$ has the value $(\quad)$
A. 2
B. 0
C. 1
D. $\frac{1}{2}$
6. C From $a_{1}=1, a_{2}=2, a_{n+2}=a_{n}+a_{n+1}, n \in \mathbf{N}_{+}$, it is easy to know that for any $n \in \mathbf{N}_{+}$, $a_{n} \in \mathbf{N}_{+}$, and $a_{1} < a_{2} < \cdots < a_{n} < \cdots$. Therefore, $a_{n+2}=a_{n}+a_{n+1}<a_{n+1}+a_{n+1}=2 a_{n+1}$, which means $\frac{a_{n+2}}{a_{n+1}}<2$, so $1<\frac{a_{n+2}}{a_{n+1}}<2$. Thus, $\left\{\frac{a_{n+2}}{a_{n+1}}\right\}=\frac{a_{n+2}}{a_{n+1}}-1=\frac{a_{n+1}+a_{n}}{a_{n+1}}-1=\frac{a_{n}}{a_{n+1}}, n \in \mathbf{N}_{+}$. Therefore, the original expression $=\left[\frac{a_{2}}{a_{1}}\right] \cdot\left[\frac{a_{98}}{a_{2}}\right] \cdot \frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdots \cdots \cdot \frac{a_{97}}{a_{98}}=\left[\frac{a_{2}}{a_{1}}\right] \cdot\left[\frac{a_{98}}{a_{2}}\right] \cdot \frac{a_{1}}{a_{98}}=2 \cdot\left[\frac{a_{98}}{2}\right] \cdot \frac{1}{a_{98}}$. From the recurrence relation $a_{n+2}=a_{n}+a_{n+1}$ and $a_{1}=1$, $a_{2}=2$, it is easy to prove by mathematical induction that $a_{3 k+2}$ is even, and $a_{3 k+1}$ and $a_{3 k+3}$ are odd $(k \in \mathbf{N})$, so $a_{98}$ is even, and the original expression $=2 \cdot \frac{a_{98}}{2} \cdot \frac{1}{a_{98}}=1$.