递推数列的特征方程，可被用于对递推数列求通项，常用于中学数学中数列的拓展教学。
给定数列满足 x n + 2 + a ⋅ x n + 1 + b ⋅ x n = 0 {\displaystyle x_{n+2}+a\cdot x_{n+1}+b\cdot x_{n}=0}
设 α, β 为 x + a ⋅ x + b = 0 {\displaystyle x+a\cdot x+b=0} 两根。
于是 x n + 2 − ( α + β ) ⋅ x n + 1 + α ⋅ α ⋅ x n = 0 x n + 2 − a ⋅ x n + 1 − β ⋅ ( x n + 1 − α ⋅ x n {\displaystyle {\begin{array}{l}x_{n+2}-(\alpha +\beta )\cdot x_{n+1}+\alpha \cdot \alpha \cdot x_{n}=0\\x_{n+2}-a\cdot x_{n+1}-\beta \cdot (x_{n+1}-\alpha \cdot x_{n}\end{array}}}
两根。
