6. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=\frac{1}{4}, a_{2}=\frac{1}{5}$, and
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}=n a_{1} a_{n+1}
$$

for any positive integer $n$. Then the value of $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{97}}$ is $(\quad)$.
(A) 5032
(B) 5044
(C) 5048
(D) 5050
6.B.

Given $a_{1} a_{2}+a_{2} a_{3}=2 a_{1} a_{3}$ and $a_{1}=\frac{1}{4}, a_{2}=\frac{1}{5}$, we get $a_{3}=\frac{1}{6}$.

Given $a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}=3 a_{1} a_{4}$, we can get $a_{4}=\frac{1}{7}$.
Furthermore, we can prove by mathematical induction that $a_{n}=\frac{1}{n+3}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{97}} \\
=4+5+\cdots+100=5044 .
\end{array}
$$