SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Note that, since $x_1=1$, $x_{n+1}=\frac{x_n+2}{x_n+1}$ we can write $$\begin{array}{rl}{x}_{n+1}& =\frac{x_n\cdot x_1+2}{x_n+x_1}=\frac{\frac{x_{n-1}+2}{x_{n-1}+1}\cdot x_1+2}{\frac{x_{n-1}+2}{x_{n-1}+1}+x_1}=\frac{x_{n-1}\left(x_1+2\right)+2\left(x_1+1\right)}{x_{n-1}\left(x_1+1\right)+\left(x_1+2\right)}\\ & =\frac{x_{n-1}\cdot \frac{x_1+2}{x_1+1}+2}{x_{n-1}+\frac{x_1+2}{x_1+1}}=\frac{x_{n-1}\cdot x_2+2}{x_{n-1}+x_2}\end{array}$$ So by the same way, we can prove that $$x_n=\frac{x_k{x}_{n-k}+2}{x_k+x_{n-k}}\text{ for any }1\le k\le n-1\text{. }$$ Let $a_{n+1}=x_{2^n}$ for $n\ge 0$ then $a_1=x_1=1$, $a_2=x_2$, $a_3=x_4$, $\dots$ Take $k=2^{n-1}$ in above formula, we get $$x_{2^n}=\frac{x_{2^{n-1}}^2+2}{2x_{2^{n-1}}}\text{ so }{a}_{n+1}=\frac{a_n^2+2}{2a_n}.$$ Thus by induction, one can get $a_n=y_n$ for all $n\ge 1$. Therefore, $y_{n+1}=a_{n+1}=x_{2^n}$ for any $n\ge 0$. $\square$