9. Let $a_{1}=1, a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)$, find the general term formula of the sequence $\left\{a_{n}\right\}$.
9. Using the substitution method, let $b_{n}=\sqrt{1+24 a_{n}}$, then $b_{1}=5$, and $a_{n}=\frac{\left(b_{n}^{2}-1\right)}{24}$. Substituting into the original recurrence relation and simplifying, we get $\left(2 b_{n+1}\right)^{2}=\left(b_{n}+3\right)^{2}$. Since $b_{n}>0$, we have $2 b_{n+1}=b_{n}+3$, which simplifies to $b_{n+1}=\frac{1}{2} b_{n}+\frac{3}{2}$. Therefore, $b_{n+1}-3=\frac{1}{2}\left(b_{n}-3\right)$. Since $b_{n}=\left(b_{1}-3\right)\left(\frac{1}{2}\right)^{n-1}+3=2^{2-n}+3$, we have $a_{n}=\frac{2^{2 n-1}+3 \cdot 2^{n-1}+1}{3 \cdot 2^{2 n-1}}(n \geqslant 1)$.