Construct a new sequence $\{b_n\}$ such that $b_n=\sqrt{1_24a_n}$. It is clear that $b_1=5$ and every term in $\{b_n\}$ is non-negative. $$b_n^2=1+24a_n\implies a_n=\frac{b_n^2-1}{24}$$ Then from the given recursion, we have $$\frac{b_{n+1}^2-1}{24}=\frac{1}{16}\Big(1+4\times\frac{b_n^2-1}{24}+b_n\Big)$$ This above relationship can be simplified to $$(2b_{n+1})^2 = (b_n+3)^2 \implies 2b_{n+1}=b_n+3\qquad(\because b_n \ge 0)$$ By the conclusion of %%HREF%%3175%%, we have $$2b_{n+1}=b_n+3 \implies b_{n}=3+\frac{1}{2^{n-2}}=3+2^{2-n}$$ $$\therefore\quad a_n=\frac{b_n^2-1}{24}=\frac{2^{2n-1}+3\times 2^{n-1}+1}{3\times 2^{2n-1}}$$