Quadratic Equations
Power Sums via Newton's Identity — Minimum Value
nta_pyq_2024_apr
Grade 11
Question:
Let $\alpha,\beta$ be the distinct roots of the equation $x^2-(t^2-5t+6)x+1=0$, $t\in\mathbb{R}$ and $a_n=\alpha^n+\beta^n$. Then the minimum value of $\dfrac{a_{2023}+a_{2025}}{a_{2024}}$ is
$-\dfrac{1}{4}$
$-\dfrac{1}{4}$
$-\dfrac{1}{2}$
$\dfrac{1}{4}$
Step-by-Step Solution
Key Concept: By the recurrence $a_{n+2}-(t^2-5t+6)a_{n+1}+a_n=0$, we get $a_{2023}+a_{2025}=(t^2-5t+6)a_{2024}$. So $\frac{a_{2023}+a_{2025}}{a_{2024}}=t^2-5t+6=\left(t-\frac{5}{2}\right)^2-\frac{1}{4}$.
$\frac{a_{2023}+a_{2025}}{a_{2024}}=t^2-5t+6$. Minimum $=-\frac{1}{4}$.
Correct Answer: 2