9. (FRG) A sequence $\left(a_{n}\right)$ is defined by means of the recursion
$$
a_{1}=1, \quad a_{n+1}=\frac{1+4 a_{n}+\sqrt{1+24 a_{n}}}{16} .
$$

Find an explicit formula for $a_{n}$.
9. If we put $1+24 a_{n}=b_{m}^{2}$, the given recurrent relation becomes
$$
\frac{2}{3} b_{n+1}^{2}=\frac{3}{2}+\frac{b_{n}^{2}}{6}+b_{n}=\frac{2}{3}\left(\frac{3}{2}+\frac{b_{n}}{2}\right)^{2} \text {, i.e., } \quad b_{n+1}=\frac{3+b_{n}}{2},
$$
where $b_{1}=5$. To solve this recurrent equation, we set $c_{n}=2^{n-1} b_{n}$. From (1) we obtain
$$
\begin{aligned}
c_{n+1} & =c_{n}+3 \cdot 2^{n-1}=\cdots=c_{1}+3\left(1+2+2^{2}+\cdots+2^{n-1}\right) \\
& =5+3\left(2^{n}-1\right)=3 \cdot 2^{n}+2 .
\end{aligned}
$$

Therefore $b_{n}=3+2^{-n+2}$ and consequently
$$
a_{n}=\frac{b_{n}^{2}-1}{24}=\frac{1}{3}\left(1+\frac{3}{2^{n}}+\frac{1}{2^{2 n-1}}\right)=\frac{1}{3}\left(1+\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^{n}}\right) .
$$