设数列 $\left\{a_{n}\right\}$ 满足 $a_{1}+3 a_{2}+\cdots+(2 n-1) a_{n}=2 n$.

(1) 求 $\left\{a_{n}\right\}$ 的通项公式;

(2) 求数列 $\left\{\frac{a_{n}}{2 n+1}\right\}$ 的前 $n$ 项和.
(1) 数列 $\left\{a_{n}\right\}$ 满足 $a_{1}+3 a_{2}+\cdots+(2 n-1) a_{n}=2 n$. $n \geq 2$ 时, $a_{1}+3 a_{2}+\cdots+(2 n-3) a_{n-1}=2(n-1) \ldots \therefore$ (2n-1) $a_{n}=2 . \therefore a_{n}=\frac{2}{2 n-1}$. 当 $n=1$ 时, $a_{1}=2$,上式也成立. $\therefore a_{n}=\frac{2}{2 n-1}$.

(2) $\frac{a_{n}}{2 n+1}=\frac{2}{(2 n-1)(2 n+1)}=\frac{1}{2 n-1}-\frac{1}{2 n+1} . \therefore$ 数列 $\left\{\frac{a_{n}}{2 n+1}\right\}$ 的前 $n$ 项和 $=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{2 n-1}-\right.$ $\left.\frac{1}{2 n+1}\right)=1-\frac{1}{2 n+1}=\frac{2 n}{2 n+1}$.