For each natural number $n$, the parabola $y=\left(n^{2}+n\right) x^{2}$
$-(2 n+1) x+1$ intersects the $x$-axis at points $A_{n}, B_{n}$. Let $\left|A_{n} B_{n}\right|$ denote the distance between these two points. Then $\left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots$ $+\left|A_{1995} B_{1995}\right|$ is $\qquad$.
$$
\begin{aligned}
& \text { II.1. } 1995 \\
& \because y=(n x-1)[(n+1) x-1], \\
x_{1}= & \frac{1}{n}, x_{2}=\frac{1}{n+1}, \\
& \therefore\left|A_{n} B_{n}\right|=\frac{1}{n}-\frac{1}{n+1} .
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
& \left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots+\left|A_{1005} B_{1005}\right| \\
= & \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{1995}-\frac{1}{1996}\right) \\
=1- & \frac{1}{1996}=\frac{1995}{1996} .
\end{aligned}
$$