Example 2 In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)$, try to find the general term formula $a_{n}$.
Solution: Let $b_{n}=\sqrt{1+24 a_{n}}$, then $b_{1}=5, b_{n}^{2}=1+24 a_{n}$, substituting into the recurrence relation we get
$\frac{b_{n+1}^{2}-1}{24}=\frac{1}{16}\left(1+4 \cdot \frac{b_{n}^{2}-1}{24}+b_{n}\right), \quad$ which simplifies to $\left(2 b_{n+1}\right)^{2}=\left(b_{n}+3\right)^{2}$.
Since $b_{n}>0$, we have $2 b_{n+1}=b_{n}+3$. Let $c_{n}=b_{n}-3, c_{1}=3$, then $2 c_{n+1}$ $=2 b_{n+1}-6=b_{n}-3=c_{n}$, hence $\left\{c_{n}\right\}$ is a geometric sequence with the first term 2 and common ratio $\frac{1}{2}$, i.e., $c_{n}=2\left(\frac{1}{2}\right)^{n-1}=2^{2-n}$, from which we deduce
$$
b_{n}=c_{n}+3=3+2^{2 n-n}, a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)=\frac{2^{2 n-1}+3 \cdot 2^{n-1}+1}{3 \cdot 2^{2 n-1}} .
$$