8、Let $2 n$ real numbers $a_{1}, a_{2}, \mathrm{~L}, a_{2 n}$ satisfy the condition $\sum_{i=1}^{2 n-1}\left(a_{i+1}-a_{i}\right)^{2}=1$ then $\mu=\left(a_{n+1}+a_{n+2}+\mathrm{L}+a_{2 n}\right)-\left(a_{1}+a_{2}+\mathrm{L}+a_{n}\right)$ the maximum value is $\qquad$
8. $\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}}$

Analysis: When $n \geqslant 2$, let $x_{1}=a_{1}, x_{i+1}=a_{i+1}-a_{i} \quad i=1,2,3, \mathrm{~L}, 2 n-1$
then $\sum_{i=2}^{2 n-1} x_{i}^{2}=1, a_{i}=x_{1}+x_{2}+\mathrm{L}+x_{i}$, so:
$$
\begin{array}{l}
\mu=n\left(x_{1}+x_{2}+\mathrm{L}+x_{n}\right)+n x_{n+1}+(n-1) x_{n-2}+\mathrm{L}+x_{2 n}-\left(n x_{1}+(n-1) x_{2}+\mathrm{L}+x_{n}\right) \\
=x_{2}+2 x_{3}+\mathrm{L}+(n-1) x_{n}+n x_{n+1}+(n-1) x_{n+2}+\mathrm{L}+x_{2 n} \\
\leqslant \sqrt{\left(1^{2}+2^{2}+\mathrm{L}+n^{2}+\mathrm{L}+1^{2}\right) \sum_{i=2}^{2 n-1} x_{i}^{2}}=\sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} \\
\end{array}
$$