1. When $n=1,2, \cdots, 2006$, the sum of the lengths of the segments cut off by the x-axis for all quadratic functions $y=n(n+1) x^{2}-(2 n+1) x+1$ is $\qquad$ .
2. 1. $\frac{2006}{2007}$.
Let $y=0$, we get $n(n+1) x^{2}-(2 n+1) x+1=0$.
Transform it into $(n x-1)[(n+1) x-1]=0$, we get
$x_{1}=\frac{1}{n+1}, x_{2}=\frac{1}{n}$.
Thus, $\left|x_{2}-x_{1}\right|=\frac{1}{n}-\frac{1}{n+1}$.
By taking $n=1,2, \cdots, 2006$, we get the sum of the lengths of the segments intercepted on the $x$-axis by these 2006 parabolas as
$$
\begin{array}{l}
S_{2006}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{2006}-\frac{1}{2007}\right) \\
=1-\frac{1}{2007}=\frac{2006}{2007} .
\end{array}
$$