8.26 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, n \geqslant 1 .
$$

Find the general term formula for $a_{n}$.
[Solution] When $n \geqslant 2$, since
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n-1}=(n-1)^{2} a_{n-1},
\end{array}
$$

Therefore,
From this, we can get
Thus, we obtain
$$
\begin{array}{l}
a_{n}=n^{2} a_{n}-(n-1)^{2} a_{n-1} . \\
a_{n}=\frac{n-1}{n+1} a_{n-1}, n=2,3, \cdots \\
a_{n}=\frac{1}{n(n+1)}, n=1,2,3, \cdots
\end{array}
$$