amparvardi wrote: A sequence $(a_n)$ is defined by means of the recursion \[a_1 = 1, a_{n+1} = \frac{1 + 4a_n +\sqrt{1+ 24a_n}}{16}.\] Find an explicit formula for $a_n.$ So $24a_{n+1}+1=\frac {9}4+\frac{24a_n+1}4+\frac 32\sqrt{24a_n+1}$ $=\left(\frac{\sqrt{24a_n+1}+3}2\right)^2$ Let then $b_n=\sqrt{24a_n+1}$. We get $b_1=5$ and $b_{n+1}=\frac{b_n+3}2$ and so $b_{n+1}-3=\frac{b_n-3}2$ So $b_n=3+2^{2-n}$ and $\boxed{a_n=\frac{(3+2^{2-n})^2-1}{24}}$