6. Given the quadratic function $y=x^{2}-\frac{2 n+1}{n(n+2)} x+\frac{n+1}{n(n+2)^{2}}$ which intercepts a segment of length $d_{n}$ on the $x$-axis, then $\sum_{n=1}^{100} d_{n}=$ $\qquad$ .
6 $\frac{7625}{10302}$ Hint: Since
$$
d_{n}=\sqrt{\left(\frac{2 n+1}{n(n+2)}\right)^{2}-\frac{4(n+1)}{n(n+2)^{2}}}=\frac{1}{n(n+2)},
$$

thus
$$
\begin{aligned}
\sum_{n=1}^{100} d_{n} & =\frac{1}{2} \sum_{n=1}^{100}\left(\frac{1}{n}-\frac{1}{n+2}\right) \\
& =\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{101}-\frac{1}{102}\right)=\frac{7625}{10302} .
\end{aligned}
$$