Let [, ] be the distinct roots of the equation [x^2-(t^2-5 t+6) x+1=0, t R] and [a_n=^n+^n…

Question

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

Accepted Answer

$$\begin{aligned}
& x^2-\left(t^2-5 t+6\right) x+1=0 \
& \therefore a_{2025}-\left(t^2-5 t+6\right) a_{2024}+a_{2023}=0 \
& \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \
& =\left(t+\frac{5}{2}\right)^2+\left(\frac{-1}{4}\right) \
& \text { Minimum value }=\frac{-1}{4}
\end{aligned}$$

Solutions · 1