Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product of the roots is $c/a$.
• Characteristic Equation: Given the roots of a quadratic equation, the quadratic can be written as $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
• Recurrence Relations: A recurrence relation defines a sequence where each term is a function of the preceding terms.

Step-by-Step Solution

1. Formulating the Characteristic Quadratic Equation

• Explanation: We are given $\alpha + \beta = 1$ and $\alpha\beta = -1$. We will use Vieta's formulas to construct the quadratic equation whose roots are $\alpha$ and $\beta$.
• Working: The quadratic equation is given by $x^2 - (\alpha + \beta)x + \alpha\beta = 0$. Substituting the given values, we get: $x^2 - (1)x + (-1) = 0$ $x^2 - x - 1 = 0$
• Why: This quadratic equation is crucial because $\alpha$ and $\beta$ satisfy it.

2. Deriving the Recurrence Relation for $p_n$

• Explanation: Since $\alpha$ and $\beta$ are roots of $x^2 - x - 1 = 0$, we have $\alpha^2 - \alpha - 1 = 0$ and $\beta^2 - \beta - 1 = 0$. This can be rewritten as $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$. We want to find a recurrence relation for $p_n = \alpha^n + \beta^n$.
• Working: Multiplying $\alpha^2 = \alpha + 1$ by $\alpha^{n-1}$, we get $\alpha^{n+1} = \alpha^n + \alpha^{n-1}$ Similarly, multiplying $\beta^2 = \beta + 1$ by $\beta^{n-1}$, we get $\beta^{n+1} = \beta^n + \beta^{n-1}$ Adding these two equations, we obtain $\alpha^{n+1} + \beta^{n+1} = (\alpha^n + \beta^n) + (\alpha^{n-1} + \beta^{n-1})$
• Why: This step links the sum of powers at index $n+1$ to the sums of powers at indices $n$ and $n-1$. Using the definition $p_k = \alpha^k + \beta^k$, we can rewrite the equation as: $p_{n+1} = p_n + p_{n-1}$
• Why: This is the recurrence relation we need.

3. Solving for $p_n$

• Explanation: We are given that $p_{n-1} = 11$ and $p_{n+1} = 29$. We will use the recurrence relation to solve for $p_n$.
• Working: We have $p_{n+1} = p_n + p_{n-1}$. Substituting the given values, we get $29 = p_n + 11$ $p_n = 29 - 11$ $p_n = 18$

4. Using the initial conditions to solve for n and alpha, beta

• Explanation: We need to use the given initial conditions to solve for $p_n^2$.
• Working: Since $p_n = \alpha^n + \beta^n$, we have $p_{n-1} = \alpha^{n-1} + \beta^{n-1} = 11$ and $p_{n+1} = \alpha^{n+1} + \beta^{n+1} = 29$. We also found that $p_n = \alpha^n + \beta^n = 18$. We have $\alpha + \beta = 1$ and $\alpha \beta = -1$. Therefore, $\alpha = \frac{1 + \sqrt{5}}{2}$ and $\beta = \frac{1 - \sqrt{5}}{2}$ or vice-versa.

5. Finding $p_1$ and $p_2$

• Explanation: We want to see if we can find values for $p_1$ and $p_2$.
• Working: Since $p_1 = \alpha^1 + \beta^1 = \alpha + \beta = 1$ and $p_2 = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (1)^2 - 2(-1) = 1 + 2 = 3$.

6. Finding n

• Explanation: Since we have the recurrence relation, let's try to find n using the first few values.
• Working: $p_1 = 1$ $p_2 = 3$ $p_3 = p_2 + p_1 = 3 + 1 = 4$ $p_4 = p_3 + p_2 = 4 + 3 = 7$ $p_5 = p_4 + p_3 = 7 + 4 = 11$ $p_6 = p_5 + p_4 = 11 + 7 = 18$ $p_7 = p_6 + p_5 = 18 + 11 = 29$ Therefore, $p_{n-1} = 11, p_n = 18, p_{n+1} = 29$. We have $n-1 = 5$, $n = 6$, and $n+1 = 7$.

7. Calculating $p_n^2$

• Explanation: The problem asks for the value of $p_n^2$. We have found that $p_n = 18$. However, the correct answer is 324. The correct answer given is 1.

8. Recalculating

• Explanation: There must be an error. $p_n = 18$, so something is wrong.
• Working: $p_{n+1} = p_n + p_{n-1}$ If $p_n = \alpha^n + \beta^n$ $p_0 = \alpha^0 + \beta^0 = 1 + 1 = 2$ $p_1 = \alpha + \beta = 1$ $p_2 = 3$ $p_3 = 4$ $p_4 = 7$ $p_5 = 11$ $p_6 = 18$ $p_7 = 29$

We are given $p_{n-1} = 11$ and $p_{n+1} = 29$. This means that $n-1=5$ and $n+1=7$. Thus $n=6$. $p_6 = 18$. The question is asking for a number $p_n^2$. The correct answer is 1.

$p_{n+1}p_{n-1} - p_n^2 = (\alpha^{n+1} + \beta^{n+1})(\alpha^{n-1} + \beta^{n-1}) - (\alpha^n + \beta^n)^2 = \alpha^{2n} + \alpha^{n+1}\beta^{n-1} + \beta^{n+1}\alpha^{n-1} + \beta^{2n} - (\alpha^{2n} + 2\alpha^n\beta^n + \beta^{2n}) = \alpha^{n-1}\beta^{n-1}(\alpha^2 + \beta^2) - 2\alpha^n\beta^n = (\alpha\beta)^{n-1}(\alpha^2 + \beta^2) - 2(\alpha\beta)^n = (-1)^{n-1}(3) - 2(-1)^n = (-1)^{n-1}[3 + 2(-1)] = (-1)^{n-1}[3 - 2] = (-1)^{n-1}$ Since $n=6$, $p_{n+1}p_{n-1} - p_n^2 = (-1)^{6-1} = -1$. $p_{n+1}p_{n-1} = 29(11) = 319$. $p_n^2 = 319 - (-1) = 320$.

Let $q_n = p_{n+1}p_{n-1} - p_n^2$ $q_6 = p_7p_5 - p_6^2 = 29(11) - 18^2 = 319 - 324 = -5$. Since $\alpha\beta = -1$, $p_{n+1}p_{n-1} - p_n^2 = -(\alpha\beta)^{n-1}(\alpha - \beta)^2$. Since $\alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{1 + 4} = \sqrt{5}$. Then $p_{n+1}p_{n-1} - p_n^2 = -(-1)^{n-1}(\sqrt{5})^2 = -5(-1)^{n-1}$.

The question asks for the value of $p_n^2$, but the correct answer is 1. Let's assume the question is $p_{n+1}p_{n-1} - p_n^2$. Then it will be equal to $5$ if $n$ is even and $-5$ if $n$ is odd. However, neither matches.

If $\alpha+\beta=1, \alpha\beta=-1$, then $\alpha^2 - \alpha - 1 = 0$. $p_{n+2} = \alpha^{n+2} + \beta^{n+2} = \alpha^{n}(\alpha^2) + \beta^{n}(\beta^2) = \alpha^{n}(\alpha + 1) + \beta^{n}(\beta+1) = \alpha^{n+1} + \alpha^{n} + \beta^{n+1} + \beta^{n} = p_{n+1} + p_n$. Therefore, $p_{n+2} = p_{n+1} + p_n$. Then $p_{n+2}p_n - p_{n+1}^2 = (\alpha^{n+2} + \beta^{n+2})(\alpha^n + \beta^n) - (\alpha^{n+1} + \beta^{n+1})^2 = -(\alpha\beta)^n(\alpha^2 + \beta^2 - 2\alpha\beta) = -(\alpha\beta)^n(\alpha-\beta)^2 = -(-1)^n((\alpha+\beta)^2 - 4\alpha\beta) = -(-1)^n(1 + 4) = -5(-1)^n$. If $p_{n+1} = 29$, then $p_n$ is not necessarily 18.

Common Mistakes & Tips

• Carefully check the arithmetic when deriving the recurrence relation and solving for $p_n$.
• Be mindful of the signs when manipulating the equations.
• If the given answer doesn't match the derived result, re-examine the steps and the problem statement.

Summary

We derived the recurrence relation $p_{n+1} = p_n + p_{n-1}$ and used the given values $p_{n-1} = 11$ and $p_{n+1} = 29$ to find that $p_n = 18$. The problem asks for $p_n^2$, which is $18^2 = 324$. However, the problem states that the answer is 1. The question must be incorrect. The final answer is 324. The problem asks for the value of $p_n^2$, which should be 324. The answer is incorrect. Let's assume the question is $p_{n+1}p_{n-1} - p_n^2$. Given $p_{n-1} = 11$ and $p_{n+1} = 29$, then $p_n = 18$. $29(11) - 18^2 = 319 - 324 = -5$.

If instead the question is $p_n p_{n+2} - p_{n+1}^2$, we would get $11 \cdot 18 = 198 - 29^2 = 841 = -841+198 = -643$.

Final Answer

The final answer is \boxed{324}.