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 $\sum_{i=1}^{1992}\left|A_{i} B_{i}\right|$ is
A. $\frac{1991}{1992}$
B. $\frac{1992}{1993}$
C. $\frac{1991}{1993}$
D. $\frac{1993}{1992}$
$$
\left|A_{n} B_{n}\right|=\frac{\sqrt{(2 n+1)^{2}-4\left(n^{2}+n\right)}}{n^{2}+n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \text {, }
$$

Thus, $\sum_{i=1}^{1992}\left|A_{i} B_{i}\right|=\sum_{i=1}^{1992}\left(\frac{1}{i}-\frac{1}{i+1}\right)=1-\frac{1}{1993}=\frac{1992}{1993}$, so the answer is $B$.