Example 1 Given in the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, and $a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)$.
Find $a_{n}$. 1$]$

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Let $b_{n}=\sqrt{1+24 a_{n}} \geqslant 0$. Then $a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)$.
Thus, $4 b_{n+1}^{2}=\left(b_{n}+3\right)^{2}$.
Hence $2 b_{n+1}=b_{n}+3 \Rightarrow \frac{b_{n+1}-3}{b_{n}-3}=\frac{1}{2}$.
It is easy to get $b_{n}=3+\left(\frac{1}{2}\right)^{n-2}$.
Therefore, $a_{n}=\frac{1}{24}\left[\left(\frac{1}{2}\right)^{n-2}+3\right]^{2}-\frac{1}{24}$.