The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=1, a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)$, find its general term formula.
Analysis: The part with the square root is difficult to handle, and squaring it would lead to a more complicated relationship, which suggests making a substitution: let $b_{n}=\sqrt{1+24 a_{n}}$.

Solution: Let $b_{n}=\sqrt{1+24 a_{n}}$, then $a_{n}=\frac{b_{n}^{2}-1}{24}, b_{1}=5$. Thus, the original recurrence relation can be transformed into $\frac{b_{n+1}^{2}-1}{24}=\frac{1}{16}\left(1+4 \cdot \frac{b_{n}^{2}-1}{24}+b_{n}\right)$, which simplifies to $\left(2 b_{n+1}\right)^{2}=\left(b_{n}+3\right)^{2}$. Since $b_{n}, b_{n+1}$ are non-negative, we have $2 b_{n+1}=b_{n}+3$, hence $b_{n+1}-3=\frac{1}{2}\left(b_{n}-3\right)$. Therefore, $b_{n+1}-3=\left(b_{n}-3\right)\left(\frac{1}{2}\right)^{n-2}$, which means $b_{n}=3+\left(\frac{1}{2}\right)^{n-2}$. Thus, $a_{n}=\frac{b_{n}^{2}-1}{24}=\frac{1}{3}+\frac{1}{3 \cdot 2^{2 n-1}}+\frac{1}{2^{n}}$.