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

According to Newton's theorem, the recurrence relation for \( a_n \) is:

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

Applying this relation:

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

Substituting into the expression:

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

The quadratic \( t^2 - 5t + 6 \) can be rewritten 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 \( 0 \), achieved when \( t = \frac{5}{2} \). Substituting this into the equation yields:

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