11. The sequence $x_{1}=0, x_{n+1}=\frac{n+2}{n} x_{n}+\frac{1}{n}$, the general term formula of the sequence $\left\{a_{n}\right\}$ is
11. $x_{n}=\frac{1}{4}(n-1)(n+2)$.
$x_{n+1}=\frac{n+2}{n} x_{n}+\frac{1}{n}$, adding $\frac{1}{2}$ to both sides gives $\left(x_{n+1}+\frac{1}{2}\right)=\frac{n+2}{n}\left(x_{n}+\frac{1}{2}\right)$, which can be transformed to $\frac{1}{(n+2)(n+1)}\left(x_{n+1}+\frac{1}{2}\right)=\frac{1}{(n+1) n}\left(x_{n}+\frac{1}{2}\right)=\cdots=\frac{1}{2-1}\left(x_{1}+\frac{1}{2}\right)=\frac{1}{4}$, hence we have $x_{n}+$ $\frac{1}{2}=\frac{1}{4}(n+1) n$. Therefore, $x_{n}=\frac{1}{4}(n+2)(n-1)\left(n \in \mathbf{N}_{+}\right)$.