4. The sequence $\left\{a_{n}\right\}$ is defined by the formula $\left\{\begin{array}{l}a_{1}=1 \\ a_{n+1}=\frac{1}{16}\left(1+4 a_{n}+\sqrt{1+24 a_{n}}\right)\end{array}, n=1,2, \cdots\right.$, find the general term of $\left\{a_{n}\right\}$.
4. Solution Let $b_{n}=\sqrt{1+24 a_{n}}$, then $b_{1}=5, b_{n}^{2}=1+24 a_{n}$, substituting $a_{n}=\frac{b_{n}^{2}-1}{24}$ into the original equation, we get $\frac{b_{n-1}^{2}-1}{24}=$ $\frac{1}{16}\left(1+4 \cdot \frac{b_{n}^{2}-1}{24}+b_{n}\right), \therefore\left(2 b_{n-1}\right)^{2}=\left(b_{n}+3\right)^{2}, \because b_{n}>0, \therefore 2 b_{n-1}=b_{n}+3$, solving by the method of undetermined coefficients, we get $b_{n}=$ $3+\left(\frac{1}{2}\right)^{n-2}$
$$
\therefore a_{n}=\frac{b_{n}^{2}-1}{24}=\frac{1}{3}+\left(\frac{1}{2}\right)^{n}+\frac{2}{3}\left(\frac{1}{4}\right)^{n} .
$$