Example 4 (22nd IMO Preliminary) In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)$. Find $a_{n}$.
Solve: Construct a new sequence $\left\{b_{n}\right\}$, such that $b_{n}=\sqrt{1+24 a_{n}}>0$, then $b_{1}=5, b_{n}^{2}=1+24 a_{n}$, i.e., $a_{n}=\frac{b_{n}^{2}-1}{24}$.
$$
\therefore \frac{b_{n+1}^{2}-1}{24}=\frac{1}{16}\left(1+4 \times \frac{b_{n}^{2}-1}{24}+b_{n}\right) \text {, simplify }
$$

to get $\left(2 b_{n+1}\right)^{2}=\left(b_{n}+3\right)^{2}, \therefore 2 b_{n+1}=b_{n}+3$,
i.e., $b_{n+1}-3=\frac{1}{2}\left(b_{n}-3\right)$, then the sequence $\left\{b_{n}-3\right\}$ is a geometric sequence with 2 as the first term and $\frac{1}{2}$ as the common ratio.
$$
\begin{array}{l} 
b_{n}-3=2 \times\left(\frac{1}{2}\right)^{n-1}=2^{2-n}, \text { i.e., } b_{n}=2^{2-n}+ \\
3, \therefore a_{n}=\frac{b_{n}^{2}-1}{24}=\frac{2^{2 n-1}+3 \times 2^{n-1}+1}{3 \times 2^{2 n-1}} .
\end{array}
$$