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

Question

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

Accepted Answer

$\begin{gathered} {x}^{2}-({t}^{2}-5t+6)x+1=0 \ 	herefore {a}_{2025}-({t}^{2}-5t+6){a}_{2024}+{a}_{2023}=0 \ \Rightarrow \frac{{a}_{2025}+{a}_{2023}}{{a}_{2024}}={t}^{2}-5t+6 \ ={(t+\frac{5}{2})}^{2}+(\frac{-1}{4}) \ =\frac{-1}{4} \end{gathered}$

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