Example 13 Given $a_{1}=1, a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)(n \geqslant 1)$, find $a_{n}$.
(1986 Federal Republic of Germany Competition Problem)
Let $b_{n}=\sqrt{1+24 a_{n}}$, i.e., $a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)$, the original recurrence relation becomes $\left(2 b_{n+1}+1\right)^{2}=\left(b_{n}+3\right)^{2}$.
Since $b_{n}>0$, we have $2 b_{n+1}=b_{n}+3$, i.e., $b_{n+1}=\frac{1}{2} b_{n}+\frac{3}{2}$, and $b_{1}=\sqrt{1+24 a_{1}}=5$. The fixed point of $f(x)=\frac{1}{2} x+\frac{3}{2}$ is 3, so $b_{n}=\left(\frac{1}{2}\right)^{n-1}\left(b_{1}-3\right)+3=2^{2-n}+3$, hence $a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)=\left(2^{2 n-1}+3 \cdot 2^{n-1}+1\right) /\left(3 \cdot 2^{2 n-1}\right), n \geqslant 1$.