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:
- \( \frac{1}{4} \)
- \( -\frac{1}{2} \)
- \( -\frac{1}{4} \)
- \( \frac{1}{2} \)
The Correct Option is C
Approach Solution - 1
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:
\[ \text{Minimum value} = -\frac{1}{4}. \]
Approach Solution -2
To solve the problem, we need to find the minimum value of \( \frac{a_{2023} + a_{2025}}{a_{2024}} \), where \( a_n = \alpha^n + \beta^n \) and \( \alpha, \beta \) are roots of the quadratic equation \( x^2 - (t^2 - 5t + 6)x + 1 = 0 \).
First, let's analyze the equation \( x^2 - (t^2 - 5t + 6)x + 1 = 0 \). The roots \( \alpha \) and \( \beta \) are given by the equation:
\(\alpha + \beta = t^2 - 5t + 6\)
\(\alpha \beta = 1\)
Using the property of roots of quadratic equations, the sequence \( a_n \) can be derived from the recurrence relation:
\(a_n = (t^2 - 5t + 6) a_{n-1} - a_{n-2}\)
Given initial conditions for sequence:
\(a_0 = 2\) (since \( a_0 = \alpha^0 + \beta^0 = 1 + 1 \))
\(a_1 = t^2 - 5t + 6\) (since \( a_1 = \alpha + \beta \))
The required expression is:
\(\frac{a_{2023} + a_{2025}}{a_{2024}}\)
This expression is equivalent to:
\(a_{2024} \, = \, (t^2 - 5t + 6) a_{2023} - a_{2022}\)
\(a_{2025} \, = \, (t^2 - 5t + 6) a_{2024} - a_{2023}\)
Thus, the expression becomes:
\(\frac{a_{2023} + ((t^2 - 5t + 6) a_{2024} - a_{2023})}{a_{2024}} \, = \, \frac{(t^2 - 5t + 6) a_{2024}}{a_{2024}}\)
Simplifying this, we find:
\(\frac{a_{2023} + a_{2025}}{a_{2024}} = t^2 - 5t + 6 - 2\)
Which further simplifies to:
\(t^2 - 5t + 4\)
To find the minimum value of \( t^2 - 5t + 4 \), we complete the square:
\(t^2 - 5t + 4 = (t - \frac{5}{2})^2 - \frac{9}{4}\)
The minimum value occurs at \( t = \frac{5}{2} \), with the minimum value being:
\(-\frac{9}{4}\)
Therefore, the minimum value of \( \frac{a_{2023} + a_{2025}}{a_{2024}} \) is \(-\frac{1}{4}\).
Hence, the correct answer is \(-\frac{1}{4}\).
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