Example 7 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 the general term formula.
(1986 Federal Republic of Germany Olympiad Problem)
Let $b_{n}=\sqrt{1+24 a_{n}}$, then for $n>1$, from the recursive relation of the sequence $\left\{a_{n}\right\}$, we have
$$
\begin{aligned}
b_{n}^{2} & =1+24 a_{n}=1+\frac{24}{16}\left(1+4 a_{n-1}+\sqrt{1+24 a_{n-1}}\right) \\
& =\frac{1}{4}\left(10+24 a_{n-1}+6 \sqrt{1+24 a_{n-1}}\right) \\
& =\frac{1}{4}\left(\sqrt{1+24 a_{n-1}}+3\right)^{2} \\
& =\frac{1}{4}\left(3+b_{n-1}\right)^{2} .
\end{aligned}
$$

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