Let \alpha, \beta be the distinct roots of the equation x^2-\left(t^2-5 t+6\right) x+1=0, t \in \mathbb{R} and a_n=\alpha^n+\beta^n. Then the minimum value of \frac{a_{2023}+a_{2025}}{a_{2024}} is

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Answer: B. -1 / 4 \begin{aligned} & x^2-\left(t^2-5 t+6ight) x+1=0 \ & 	herefore a_{2025}-\left(t^2-5 t+6ight) a_{2024}+a_{2023}=0 \ & \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \ & =\left(t+\frac{5}{2}ight)^2+\left(\frac{-1}{4}ight) \ & 	ext { Minimum value }=\frac{-1}{4} \end{aligned}

Let $$\alpha, \beta$$ be the distinct roots of the equation $$x^2-\left(t^2-5 t+6\right) x+1=0, t \in \mathbb{R}$$ and $$a_n=\alpha^n+\beta^n$$. Then the minimum value of $$\frac{a_{2023}+a_{2025}}{a_{2024}}$$ is

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