Recurrence Relation for Powers of Roots

For the quadratic
$$x^{2} - (t^{2}-5t+6)x + 1 = 0$$
let the two distinct roots be $$\alpha$$ and $$\beta$$.

By Vieta’s relations
$$\alpha + \beta = t^{2}-5t+6 \quad -(1)$$
$$\alpha\beta = 1 \quad -(2)$$

Define the sequence $$a_{n} = \alpha^{n} + \beta^{n}\;(n \ge 0).$$

Case 1: Recurrence for $$a_n$$
Using $$\alpha\beta = 1$$ we obtain a standard identity: for all $$n \ge 0$$
$$\alpha^{\,n+2} + \beta^{\,n+2} = (\alpha+\beta)\,(\alpha^{\,n+1} + \beta^{\,n+1}) - (\alpha\beta)\,(\alpha^{\,n} + \beta^{\,n}).$$
Hence
$$a_{\,n+2} = (\alpha+\beta)\,a_{\,n+1} - a_{\,n}\quad -(3)$$

Case 2: Evaluating the required ratio
Set $$n = 2023$$ in the recurrence $$-(3)$$:
$$a_{2025} = (\alpha+\beta)\,a_{2024} - a_{2023}\quad -(4)$$

Add $$a_{2023}$$ to both sides of $$-(4)$$:
$$a_{2023} + a_{2025} = (\alpha+\beta)\,a_{2024}$$

Therefore
$$\frac{a_{2023} + a_{2025}}{a_{2024}} = \alpha + \beta$$

From $$-(1)$$ the ratio equals
$$\alpha + \beta = t^{2} - 5t + 6\quad -(5)$$

Case 3: Minimising the expression
We must find the minimum of the quadratic function
$$f(t) = t^{2} - 5t + 6$$
for real $$t$$, under the condition that the roots $$\alpha, \beta$$ are distinct.

Write $$f(t)$$ in completed-square form:
$$f(t) = \bigl(t - \tfrac{5}{2}\bigr)^{2} - \tfrac{1}{4}$$

The square term is always non-negative, so the minimum value occurs at $$t = \tfrac{5}{2}$$ and equals
$$f_{\min} = -\tfrac{1}{4}$$

Case 4: Checking the conditions
(i) Distinct roots: the discriminant of the given quadratic in $$x$$ is
$$\Delta = (\alpha+\beta)^{2} - 4\alpha\beta = f(t)^{2} - 4$$

At $$t = \tfrac{5}{2}$$, $$f(t) = -\tfrac14$$, so
$$\Delta = \bigl(-\tfrac14\bigr)^{2} - 4 = \tfrac1{16} - 4 = -\tfrac{63}{16} \neq 0.$$
Thus the roots are indeed distinct (they form a non-real conjugate pair).

(ii) Denominator non-zero: with distinct roots, $$a_{2024} = \alpha^{2024} + \beta^{2024}$$ cannot vanish for all real $$t$$; in particular it is non-zero at $$t = \tfrac{5}{2}$$, so the ratio is well defined at the minimum.

Hence the minimum possible value of $$\dfrac{a_{2023}+a_{2025}}{a_{2024}}$$ is
$$-\tfrac14.$$

Option B is correct.

We start from the quadratic equation $$x^{2}+5\sqrt{2}\,x+10=0$$ whose roots are given to be $$\alpha$$ and $$\beta$$ with $$\alpha >\beta$$. From the standard relation between coefficients and roots of a quadratic, we immediately obtain

$$\alpha+\beta=-5\sqrt{2},\qquad \alpha\beta=10.$$

For every positive integer $$n$$ we define $$P_{n}=\alpha^{n}-\beta^{n}$$ and wish to evaluate

$$E=\dfrac{P_{17}P_{20}+5\sqrt{2}\,P_{17}P_{19}}{P_{18}P_{19}+5\sqrt{2}\,P_{18}^{2}}.$$

Because both $$\alpha$$ and $$\beta$$ satisfy the quadratic, we may replace $$x$$ by either root in the equation $$x^{2}+5\sqrt{2}\,x+10=0$$. Writing the result for an arbitrary positive power will give us a recurrence for the sequence $$P_{n}$$.

First, for any root $$r\in\{\alpha,\beta\}$$ we have $$r^{2}=-5\sqrt{2}\,r-10.$$ Multiplying both sides by $$r^{\,n-2}$$ (where $$n\ge 2$$) gives

$$r^{n}= -5\sqrt{2}\,r^{\,n-1}-10\,r^{\,n-2}.$$

Since this equality holds separately for $$r=\alpha$$ and for $$r=\beta$$, we can write the two identities and subtract them. Thus, for all integers $$n\ge 2$$,

$$\alpha^{n}-\beta^{n}= -5\sqrt{2}\left(\alpha^{\,n-1}-\beta^{\,n-1}\right)-10\left(\alpha^{\,n-2}-\beta^{\,n-2}\right).$$

Recognizing the differences as the $$P$$ sequence, we arrive at the linear recurrence relation

$$P_{n}=-5\sqrt{2}\,P_{n-1}-10\,P_{n-2}\qquad (n\ge 2).$$

Now we tackle the pieces that appear in the expression $$E$$.

First, apply the recurrence with $$n=20$$:

$$P_{20}=-5\sqrt{2}\,P_{19}-10\,P_{18}.$$

Add $$5\sqrt{2}\,P_{19}$$ to both sides to create the combination seen in the numerator:

$$P_{20}+5\sqrt{2}\,P_{19}=(-5\sqrt{2}\,P_{19}-10\,P_{18})+5\sqrt{2}\,P_{19}=-10\,P_{18}.$$

Hence the entire numerator becomes

$$P_{17}\bigl(P_{20}+5\sqrt{2}\,P_{19}\bigr)=P_{17}\times(-10\,P_{18})=-10\,P_{17}P_{18}.$$

Next, apply the recurrence with $$n=19$$ to prepare the denominator:

$$P_{19}=-5\sqrt{2}\,P_{18}-10\,P_{17}.$$

Add $$5\sqrt{2}\,P_{18}$$ to both sides:

$$P_{19}+5\sqrt{2}\,P_{18}=(-5\sqrt{2}\,P_{18}-10\,P_{17})+5\sqrt{2}\,P_{18}=-10\,P_{17}.$$

Hence the entire denominator is

$$P_{18}\bigl(P_{19}+5\sqrt{2}\,P_{18}\bigr)=P_{18}\times(-10\,P_{17})=-10\,P_{17}P_{18}.$$

Now both numerator and denominator are equal to the same quantity $$-10\,P_{17}P_{18}$$, so when we form their ratio we get

$$E=\dfrac{-10\,P_{17}P_{18}}{-10\,P_{17}P_{18}}=1.$$

So, the answer is $$1$$.

Hence, the correct answer is Option 1.

Since $$\alpha$$ and $$\beta$$ are roots of a quadratic with $$\alpha + \beta = 1$$ and $$\alpha\beta = -1$$, they satisfy $$t^2 - t - 1 = 0$$. The sequence $$p_n = \alpha^n + \beta^n$$ satisfies Newton's identity recurrence: $$p_{n+1} = (\alpha + \beta)p_n - \alpha\beta \cdot p_{n-1}$$.

Substituting the given values: $$p_{n+1} = 1 \cdot p_n - (-1) \cdot p_{n-1} = p_n + p_{n-1}$$.

We are given $$p_{n-1} = 11$$ and $$p_{n+1} = 29$$. Using the recurrence $$p_{n+1} = p_n + p_{n-1}$$, we substitute: $$29 = p_n + 11$$.

Solving for $$p_n$$: $$p_n = 29 - 11 = 18$$.

We can verify this is consistent by checking with the recurrence in the other direction. We also know $$p_0 = \alpha^0 + \beta^0 = 2$$ and $$p_1 = \alpha + \beta = 1$$. Using $$p_{k+1} = p_k + p_{k-1}$$, the sequence begins: $$p_0 = 2, p_1 = 1, p_2 = 3, p_3 = 4, p_4 = 7, p_5 = 11, p_6 = 18, p_7 = 29, \ldots$$ So $$n - 1 = 5$$, i.e., $$n = 6$$, and indeed $$p_6 = 18$$ and $$p_7 = 29$$.

Therefore, $$p_n^2 = 18^2 = 324$$.