2・104 Let $n \geqslant 2, x_{1}, x_{2}, \cdots, x_{n}$ be real numbers, and
$$\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}=1$$

For every fixed $k(k \in N, 1 \leqslant k \leqslant n)$, find the maximum value of $\left|x_{k}\right|$.
[Solution] Let $s=\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}$, then
$$\begin{aligned}
s= & \left(x_{1}+\frac{1}{2} x_{2}\right)^{2}+\frac{3}{4} x_{2}^{2}+\sum_{i=2}^{n-1} x_{i} x_{i+1}+\sum_{i=3}^{n} x_{i}^{2} \\
= & \left(x_{1}+\frac{1}{2} x_{2}\right)^{2}+\left(\sqrt{\frac{3}{4}} x_{2}+\frac{1}{2 \cdot \sqrt{\frac{3}{4}}} x_{3}\right)^{2}+\left(1-\frac{1}{4 \cdot \frac{3}{4}}\right) x_{3}^{2} \\
& +\sum_{i=3}^{n-1} x_{i} x_{i+1}+\sum_{i=4}^{n} x_{i}^{2}
\end{aligned}$$

Let $a_{1}=1, a_{2}=\frac{3}{4}, a_{i}=1-\frac{1}{4 a_{i-1}}(i=2,3, \cdots, n)$
By induction, it is easy to get
$$a_{i}=\frac{i+1}{2 i}, i=1,2, \cdots, n$$

Therefore,
$$\begin{aligned}
1- & a_{i}=1-\frac{i+1}{2 i}=\frac{i-1}{2 i} \\
s= & \left(\sqrt{a_{1}} x_{1}+\sqrt{1-a_{2}} x_{2}\right)^{2}+\left(\sqrt{a_{2}} x_{2}+\sqrt{1-a_{3}} x_{3}\right)^{2}+\cdots+ \\
& \left(\sqrt{a_{n-1}} x_{n-1}+\sqrt{1-a_{n}} x_{n}\right)^{2}+a_{n} x_{n}^{2}
\end{aligned}$$

Since $s=1$, therefore
$$\begin{array}{l}
a_{n} x_{n}^{2} \leqslant 1 \\
\left|x_{n}\right| \leqslant \sqrt{\frac{2 n}{n+1}}
\end{array}$$

It is easy to see that $\quad \max \left|x_{n}\right|=\sqrt{\frac{2 n}{n+1}}$.
Similarly,
$$\max \left|x_{1}\right|=\sqrt{\frac{2 n}{n+1}}$$

When $1<k<n$,
$$\begin{aligned}
s= & \left(\sqrt{a_{1}} x_{1}+\sqrt{1-a_{2}} x_{2}\right)^{2}+\cdots+\left(\sqrt{a_{k-1}} x_{k-1}+\sqrt{1-a_{k}} x_{k}\right)^{2}+ \\
& \left(\sqrt{a_{1}} x_{n}+\sqrt{1-a_{2}} x_{n-1}\right)^{2}+\cdots+\left(\sqrt{a_{n-k} x_{k+1}}+\right. \\
& \left.\sqrt{1-a_{n-k+1}} x_{k}\right)^{2}+\left[1-\left(1-a_{k}\right)-\left(1-a_{n-k+1}\right)\right] x_{k}^{2}
\end{aligned}$$

where
$$\begin{aligned}
& 1-\left(1-a_{k}\right)-\left(1-a_{n-k+1}\right) \\
= & a_{k}+a_{n-k+1}-1 \\
= & \frac{k+1}{2 k}+\frac{n-k+2}{2(n-k+1)}-1 \\
= & \frac{n+1}{2 k(n-k+1)}
\end{aligned}$$

Therefore,
It is easy to see that
$$\left|x_{k}\right| \leqslant \sqrt{\frac{2 k(n-k+1)}{n+1}}$$
$$\max \left|x_{k}\right|=\sqrt{\frac{2 k(n-k+1)}{n+1}}$$