11. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and $a_{n+1}=\frac{\left(n^{2}+n\right) a_{n}}{3 a_{n}+n^{2}+n}\left(n \in \mathbf{N}^{*}\right)$, then its general term formula $a_{n}=$
11. $a_{n}=\frac{n}{4 n-3}\left(n \in \mathbf{N}^{*}\right)$.

Brief solution: Taking the reciprocal of both sides gives $\frac{1}{a_{n+1}}-\frac{1}{a_{n}}=\frac{3}{n^{2}+n}=3\left(\frac{1}{n}-\frac{1}{n+1}\right)$. Therefore, $\frac{1}{a_{n+1}}-\frac{1}{a_{1}}=3\left(1-\frac{1}{n+1}\right)$. Hence, $a_{n+1}=\frac{n+1}{4 n+1}=\frac{n}{4 n-3}\left(n \in \mathbf{N}^{*}\right)$