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Mathematics Question on Sequence and series

Let α,β\alpha, \beta be the distinct roots of the equation x2−(t2−5t+6)x+1=0, t∈R and an=αn+βn. 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 a2023+a2025a2024\frac{a_{2023} + a_{2025}}{a_{2024}} is:

A

14\frac{1}{4}

B

−12-\frac{1}{2}

C

−14-\frac{1}{4}

D

12\frac{1}{2}

Answer

−14-\frac{1}{4}

Explanation

Solution

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

an+2−(t2−5t+6)an+1+an=0.a_{n+2} - \left(t^2 - 5t + 6\right)a_{n+1} + a_n = 0.

Using this relation:

a2025+a2023=(t2−5t+6)a2024.a_{2025} + a_{2023} = \left(t^2 - 5t + 6\right)a_{2024}.

Substitute into the given expression:

a2023+a2025a2024=t2−5t+6.\frac{a_{2023} + a_{2025}}{a_{2024}} = t^2 - 5t + 6.

The quadratic t2−5t+6t^2 - 5t + 6 can be expressed as:

t2−5t+6=(t−52)2−14.t^2 - 5t + 6 = \left(t - \frac{5}{2}\right)^2 - \frac{1}{4}.

The minimum value of (t−52)2\left(t - \frac{5}{2}\right)^2 is 00, which occurs when t=52t = \frac{5}{2}. Substituting this into the equation:

Minimum value=−14.\text{Minimum value} = -\frac{1}{4}.