Example 6. Let $n$ be a natural number, try to find the total length of the segments cut off on the $x$-axis by the family of parabolas $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$.

 untranslated text:
上为自然数, 试求拋物 线族 $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$ 在 $x$ 轴上所截得的线段长度的总和。

 translated text:
Let $n$ be a natural number, try to find the total length of the segments cut off on the $x$-axis by the family of parabolas $y=\left(n^{2}+n\right) x^{2}-(4 n+2) x+4$.
Solve for the x-coordinates $x_{1}, x_{2}$ of the intersection points of the parabola (family) with the $x$-axis, which are the roots of the equation $\left(n^{2}+n\right) x^{2}-(4 n+2) x+4=0$. Without loss of generality, assume $x_{2}>x_{1}$. Then $x_{2}-x_{1}=\frac{\sqrt{\triangle}}{|a|}=\frac{2}{n}-\frac{2}{n+1}$.

At this point, for $\mathrm{n} \in \mathrm{N}$, the total length of the segments intercepted on the $\mathrm{x}$-axis by the family of parabolas is
$$
\begin{aligned}
s & =\sum_{1}^{\infty}\left(\frac{2}{n}-\frac{2}{n+1}=\lim _{n \rightarrow \infty}\left[\left(\frac{2}{1}\right.\right.\right. \\
\left.-\frac{2}{2}\right) & \left.+\left(\frac{2}{2}-\frac{2}{3}\right)+\cdots+\left(\frac{2}{n}-\frac{2}{n+1}\right)\right] \\
& =\lim _{n \rightarrow \infty}\left(\frac{2}{1}-\frac{2}{n+1}\right)=2 .
\end{aligned}
$$
(Author's affiliation: Yongchun County Finance Bureau, Fujian Province)