8-29 Let the sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfy: $a_{1}=1$,
$$a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right), n=1,2, \cdots$$

Find its general term formula.
[Solution] Let $b_{n}=\sqrt{1+24} a_{n}$, then for $n>1$, by the recursive relation of the sequence $\left\{a_{n}\right\}$, we have
$$\begin{aligned}
b_{n}^{2} & =1+24 a_{n}=1+\frac{24}{16}\left(1+4 a_{n-1}+\sqrt{1+24 a_{n-1}}\right) \\
& =\frac{1}{4}\left(10+24 a_{n-1}+6 \sqrt{1+24 a_{n-1}}\right) \\
& =\frac{1}{4}\left(\sqrt{1+24 a_{n-1}}+3\right)^{2} \\
& =\frac{1}{4}\left(3+b_{n-1}\right)^{2},
\end{aligned}$$

Thus, $b_{n}=\frac{1}{2}\left(3+b_{n-1}\right)$,
which implies $b_{n}-3=\frac{1}{2}\left(b_{n-1}-3\right)$.
Given $b_{1}=\sqrt{1+24 a_{1}}=\sqrt{1+24}=5$, we have
$$b_{n}=3+\left(\frac{1}{2}\right)^{n-2}, n=1,2,3, \cdots$$

Therefore, for $n=1,2,3, \cdots$, we have
$$a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)=\frac{1}{3}+\left(\frac{1}{2}\right)^{n}+\frac{1}{3}\left(\frac{1}{2}\right)^{2 n-1}$$