遞推數列的特徵方程，可被用於對遞推數列求通項，常用於中學數學中數列的拓展教學。
給定數列滿足 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}
設 α, β 為 x + a ⋅ x + b = 0 {\displaystyle x+a\cdot x+b=0} 兩根。
於是 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}}}
兩根。
