$(a_{n+1}+k_1)(a_n+k_2)=k_3$的通项
$\displaystyle a_n=[(-\frac{m^2}{k_3})^{n-1} (\frac{1}{a_1+s}-\frac{m}{k_3+m^2})+\frac{m}{k_3+m^2}]^{-1}-s$
$\displaystyle s=\frac{k_1+k_2-\sqrt{(k_1-k_2)^2+4k_3}}{2},m=\frac{k_1-k_2-\sqrt{(k_1-k_2)^2+4k_3}}{2}$
例如:$(a_{n+1}-1)(a_n-1)=2,a_1=2$
$s=-1-\sqrt{2},m=-\sqrt{2}$
$\displaystyle a_n=\frac{4}{(-1)^n(4+3\sqrt{2})-\sqrt{2}}+1+\sqrt{2}$ |