1. 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 the value of $\left|A_{1} B_{1}\right|$ $+\left|A_{2} B_{2}\right|+\cdots+\left|A_{1992} B_{1992}\right|$ is
(A) $\frac{1991}{1992}$;
(B) $\frac{1992}{1993}$;
(C) $\frac{1991}{1993}$;
(D) $\frac{1993}{1992}$.
Answer ( )
1. (B)
$$
\because \quad y=(n x-1)[(n+1) x-1] \text {, }
$$

When $y=0$, we get
$$
\begin{aligned}
& x_{1}=\frac{1}{n}, \quad x_{2}=\frac{1}{n+1} . \\
\therefore \quad & \left|A_{n} B_{n}\right|=\frac{1}{n}-\frac{1}{n+1} . \\
\text { Therefore } & \left|A_{1} B_{1}\right|+\left|A_{2} B_{2}\right|+\cdots+\left|A_{1992} B_{1992}\right| \\
= & \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{1992}-\frac{1}{1993}\right) \\
= & 1-\frac{1}{1993} \\
= & \frac{1992}{1993} .
\end{aligned}
$$