26.22 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 $| A_{1} B_{1} \mid$ $+\left|A_{2} B_{2}\right|+\cdots+\mid A_{1992} B_{1992}$ | is
(A) $\frac{1991}{1992}$.
(B) $\frac{1992}{1993}$.
(C) $\frac{1991}{1993}$.
(D) $\frac{1993}{1992}$.
(China High School Mathematics League, 1992)
[Solution] When $y=0$, we have $(n x-1)[(n+1) x-1]=0$,
solving for $x$ gives
$$
x_{1}=\frac{1}{n}, \quad x_{2}=\frac{1}{n+1} \text {. }
$$

Therefore,
$$
\left|A_{n} B_{n}\right|=\frac{1}{n}-\frac{1}{n+1} \text {. }
$$

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