We are given values for P_n for different n. This suggests that α and β are roots of a quadratic equation. Let the quadratic equation be x² - Ax - B = 0. Then, by Vieta's formulas, α + β = A and αβ = -B. The sequence P_n = α^n + β^n satisfies the recurrence relation P_n = A P_n-1 + B P_n-2. We can use the given values to find A and B.

We are given values for P_n for different n. This suggests that α and β are roots of a quadratic equation. Let the quadratic equation be x² - Ax - B = 0. Then, by Vieta's formulas, α + β = A and αβ = -B. The sequence P_n = α^n + β^n satisfies the recurrence relation P_n = A P_n-1 + B P_n-2. We can use the given values to find A and B.

We are given P_1 = 1. Since P_1 = α^1 + β^1 = α + β, and we know that for the quadratic equation x² - Ax - B = 0, the sum of roots is A = α + β, we can directly deduce that A=1. Now, substitute A=1 into the first equation 123 = 76A + 47B. This gives 123 = 76(1) + 47B, which simplifies to 47 = 47B, so B=1.

We are given P_1 = 1. Since P_1 = α^1 + β^1 = α + β, and we know that for the quadratic equation x² - Ax - B = 0, the sum of roots is A = α + β, we can directly deduce that A=1. Now, substitute A=1 into the first equation 123 = 76A + 47B. This gives 123 = 76(1) + 47B, which simplifies to 47 = 47B, so B=1.

With A=1 and B=1, the quadratic equation whose roots are α and β is x² - Ax - B = 0, which becomes x² - 1x - 1 = 0, or simply x² - x - 1 = 0.

If a quadratic equation has roots α and β, then the quadratic equation with roots (1)/(α) and (1)/(β) can be found by replacing x with (1)/(x) in the original equation. So, substitute x = (1)/(y) into x² - x - 1 = 0. This gives ((1)/(y))² - ((1)/(y)) - 1 = 0. Multiplying by y² (assuming y ≠ 0), we get 1 - y - y² = 0, which can be rewritten as y² + y - 1 = 0.

If a quadratic equation has roots α and β, then the quadratic equation with roots (1)/(α) and (1)/(β) can be found by replacing x with (1)/(x) in the original equation. So, substitute x = (1)/(y) into x² - x - 1 = 0. This gives ((1)/(y))² - ((1)/(y)) - 1 = 0. Multiplying by y² (assuming y ≠ 0), we get 1 - y - y² = 0, which can be rewritten as y² + y - 1 = 0.

Let $P_n=\alpha^n+\beta^n,\ n\in\mathbb{N}$ . If $P_{10}=123,\ P_9=76,\ P_8=47$ and $P_1=1$ , then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:

Answer: $x^2 + x - 1 = 0$

Step-by-step solution

Step 1: Establish the recurrence relation

We are given values for $P_n$ for different $n$ . This suggests that $\alpha$ and $\beta$ are roots of a quadratic equation. Let the quadratic equation be $x^2 - Ax - B = 0$ . Then, by Vieta's formulas, $\alpha + \beta = A$ and $\alpha\beta = -B$ . The sequence $P_n = \alpha^n + \beta^n$ satisfies the recurrence relation $P_n = A P_{n-1} + B P_{n-2}$ . We can use the given values to find $A$ and $B$ .

Step 2: Formulate system of equations

Using the given values $P_{10}=123$ , $P_9=76$ , and $P_8=47$ , we can substitute them into the recurrence relation to form a system of linear equations. For $n=10$ , we have $P_{10} = A P_9 + B P_8$ , which gives $123 = 76A + 47B$ . For $n=9$ , we have $P_9 = A P_8 + B P_7$ . However, we don't have $P_7$ . Let's use $P_9 = A P_8 + B P_7$ and $P_8 = A P_7 + B P_6$ . We need to find $P_7$ and $P_6$ or use a different approach. Let's re-evaluate the system using the given values directly.

Step 3: Solve for A and B

We are given $P_1 = 1$ . Since $P_1 = \alpha^1 + \beta^1 = \alpha + \beta$ , and we know that for the quadratic equation $x^2 - Ax - B = 0$ , the sum of roots is $A = \alpha + \beta$ , we can directly deduce that $A=1$ . Now, substitute $A=1$ into the first equation $123 = 76A + 47B$ . This gives $123 = 76(1) + 47B$ , which simplifies to $47 = 47B$ , so $B=1$ .

Step 4: Form the original quadratic equation

With $A=1$ and $B=1$ , the quadratic equation whose roots are $\alpha$ and $\beta$ is $x^2 - Ax - B = 0$ , which becomes $x^2 - 1x - 1 = 0$ , or simply $x^2 - x - 1 = 0$ .

Step 5: Find the quadratic equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$

If a quadratic equation has roots $\alpha$ and $\beta$ , then the quadratic equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ can be found by replacing $x$ with $\frac{1}{x}$ in the original equation. So, substitute $x = \frac{1}{y}$ into $x^2 - x - 1 = 0$ . This gives $\left(\frac{1}{y}\right)^2 - \left(\frac{1}{y}\right) - 1 = 0$ . Multiplying by $y^2$ (assuming $y \neq 0$ ), we get $1 - y - y^2 = 0$ , which can be rewritten as $y^2 + y - 1 = 0$ .

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Let Pn=αn+βn, n∈NP_n=\alpha^n+\beta^n,\ n\in\mathbb{N}Pn​=αn+βn, n∈N. If P10=123, P9=76, P8=47P_{10}=123,\ P_9=76,\ P_8=47P10​=123, P9​=76, P8​=47 and P1=1P_1=1P1​=1, then the quadratic equation having roots 1α\frac{1}{\alpha}α1​ and 1β\frac{1}{\beta}β1​ is: