Question

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

• A. $\frac{1}{4}$
• B. $-\frac{1}{2}$
• C. $-\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

The Correct Option is C

By Newton's theorem, the recurrence relation for $a_n$ is:

$$a_{n+2} - \left(t^2 - 5t + 6\right)a_{n+1} + a_n = 0.$$

Using this relation:

$$a_{2025} + a_{2023} = \left(t^2 - 5t + 6\right)a_{2024}.$$

Substitute into the given expression:

$$\frac{a_{2023} + a_{2025}}{a_{2024}} = t^2 - 5t + 6.$$

The quadratic $t^2 - 5t + 6$ can be expressed as:

$$t^2 - 5t + 6 = \left(t - \frac{5}{2}\right)^2 - \frac{1}{4}.$$

The minimum value of $\left(t - \frac{5}{2}\right)^2$ is $0$, which occurs when $t = \frac{5}{2}$. Substituting this into the equation:

$$\text{Minimum value} = -\frac{1}{4}.$$