Example 1 (1986 Federal German Mathematical Olympiad) Given $a_{1}=1, a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\right.$ $\left.\sqrt{1+24 a_{n}}\right)(n \geqslant 1)$, find $a_{n}$.
To eliminate the square root from the recurrence relation, we naturally let $b_{n}=\sqrt{1+24 a_{n}}$, i.e., $a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)$. Substituting into the original recurrence relation, we get $\frac{1}{24}\left(b_{n+1}^{2}-1\right)=\frac{1}{16}\left[1+4 \times \frac{1}{24}\left(b_{n}^{2}-1\right)+b_{n}\right]$, which simplifies to $\left(2 b_{n+1}\right)^{2}=\left(b_{n}+3\right)$. Since $b_{n}>0$, we have $2 b_{n+1}=b_{n}+3$, or $b_{n+1}=\frac{1}{2} b_{n}+\frac{3}{2}$. Let $x=\frac{1}{2} x+\frac{3}{2}$ (i.e., $x=3$), and subtracting the two equations gives $b_{n+1}-x=\frac{1}{2}\left(b_{n}-x\right)$, or $b_{n+1}-3=\frac{1}{2}\left(b_{n}-3\right)$. It is clear that $b_{n}-3$ is a geometric sequence with the first term $b_{1}-3=\sqrt{1+24 a_{1}}-3=5-3=2$ and common ratio $\frac{1}{2}$, so
$$
b_{n}-3=2 \cdot\left(\frac{1}{2}\right)^{n-1}, b_{n}=2^{2-n}+3 \text {. }
$$

Thus, $a_{n}=\frac{1}{24}\left(b_{n}^{2}-1\right)$
$$
=\frac{1}{24}\left[\left(2^{2-n}+3\right)^{2}-1\right]=\left(2^{2 n-1}+3 \cdot 2^{n-1}+1\right) /\left(3 \cdot 2^{2 n-1}\right)\left(n \in \mathbf{N}_{+}\right) .
$$