Example 7 Let the sequence $\left\{a_{n}\right\}$ be such that $a_{1}=1, a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)$, find $a_{n}$.
Analysis: The problem involves a radical in the recursive relationship. In mathematical problem-solving methods, there are two ways to eliminate radicals: squaring and substitution. Clearly, the former is not suitable for this problem.
Solution: Let $b_{n}=\sqrt{1+24 a_{n}}$, then $b_{1}=\sqrt{1+24 a_{1}}=5, b_{n}^{2}=1+24 a_{n}$.
Thus, $a_{n}=\frac{b_{n}^{2}-1}{24} \cdots$ (1)
Substituting (1) into the given recursive formula, we get $\frac{b_{n+1}^{2}-1}{24}=\frac{1}{16}\left(1+4 \times \frac{b_{n}^{2}-1}{24}+b_{n}\right)$.
Thus, $\left(2 b_{n+1}\right)^{2}=\left(b_{n}+3\right)^{2}$.
Since $b_{n}>0$, we have $2 b_{n+1}=b_{n}+3$, which simplifies to $2\left(b_{n+1}-3\right)=b_{n}-3$.
Therefore, $\left\{b_{n}-3\right\}$ is a geometric sequence with the first term 2 and common ratio $\frac{1}{2}$.
Thus, $b_{n}-3=2 \times\left(\frac{1}{2}\right)^{n-1}=\left(\frac{1}{2}\right)^{n-2}$, hence $b_{n}=\left(\frac{1}{2}\right)^{n-2}+3$.
Therefore, $a_{n}=\frac{b_{n}^{2}-1}{24}=\frac{\left(\frac{1}{2}\right)^{2 n-4}+3 \times\left(\frac{1}{2}\right)^{n-3}+8}{24}=\frac{1+3 \times 2^{n-1}+2^{2 n-1}}{3 \times 2^{2 n-1}}\left(n \in \mathbf{N}^{*}\right)$.