Question

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 $ \alpha $ and $ \frac{1}{\beta} $ is:

• A. \( x^2 - x + 1 = 0 \)
• B. \( x^2 + x - 1 = 0 \)
• C. \( x^2 - x - 1 = 0 \)
• D. \( x^2 + x + 1 = 0 \)

Hint:

For such sequences, derive \( \alpha + \beta \) and \( \alpha \beta \) using known terms and build required equation.

Answer 1

The Correct Option is B

We are given a sequence defined by \( P_n = \alpha^n + \beta^n \), with known values \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). We need to find the quadratic equation whose roots are \( \alpha \) and \( \frac{1}{\beta} \).

Concept Used:

If \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 - Sx + P = 0 \), where \( S = \alpha + \beta \) and \( P = \alpha\beta \), then the sequence \( P_n = \alpha^n + \beta^n \) satisfies the linear recurrence relation:

\[ P_n = S \cdot P_{n-1} - P \cdot P_{n-2} \]

Furthermore, the quadratic equation with roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) can be found by substituting \( x = \frac{1}{y} \) into the original equation and simplifying.

Step-by-Step Solution:

Step 1: Determine the recurrence relation from the given values.

We are given \( P_{10} = 123 \), \( P_9 = 76 \), and \( P_8 = 47 \). Let's check the relationship between them:

\[ P_9 + P_8 = 76 + 47 = 123 \]

Since \( P_9 + P_8 = P_{10} \), we can infer that the sequence follows the recurrence relation:

\[ P_n = P_{n-1} + P_{n-2} \]

Step 2: Find the quadratic equation for which \( \alpha \) and \( \beta \) are roots.

Comparing the general recurrence \( P_n = S \cdot P_{n-1} - P \cdot P_{n-2} \) with our specific relation \( P_n = 1 \cdot P_{n-1} - (-1) \cdot P_{n-2} \), we can identify:

\[ S = \alpha + \beta = 1 \] \[ P = \alpha\beta = -1 \]

We are given \( P_1 = 1 \), which is consistent with \( S = \alpha + \beta = 1 \). 
The quadratic equation with roots \( \alpha \) and \( \beta \) is \( x^2 - Sx + P = 0 \), which is:

\[ x^2 - x - 1 = 0 \]

Step 3: Address the roots of the required equation.

The problem asks for the quadratic equation with roots \( \alpha \) and \( \frac{1}{\beta} \). From \( \alpha\beta = -1 \), we can find \( \frac{1}{\beta} = -\alpha \). Thus, the required roots are \( \alpha \) and \( -\alpha \). The sum of these roots is \( \alpha + (-\alpha) = 0 \) and the product is \( \alpha(-\alpha) = -\alpha^2 \). 

Step 4: Find the sum and product of the new roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \).

Let the new sum be \( S' \) and the new product be \( P' \).

Sum of new roots:

\[ S' = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{S}{P} \]

Product of new roots:

\[ P' = \left(\frac{1}{\alpha}\right)\left(\frac{1}{\beta}\right) = \frac{1}{\alpha\beta} = \frac{1}{P} \]

Final Computation & Result:

Step 5: Calculate \( S' \) and \( P' \) using \( S=1 \) and \( P=-1 \).

\[ S' = \frac{1}{-1} = -1 \] \[ P' = \frac{1}{-1} = -1 \]

The new quadratic equation is \( x^2 - S'x + P' = 0 \).

\[ x^2 - (-1)x + (-1) = 0 \]

\( x^2 + x - 1 = 0 \)

This matches one of the given options.

Answer 2

Given: \[ P_{10} = \alpha^{10} + \beta^{10} = 123, \quad P_9 = \alpha^9 + \beta^9 = 76, \quad P_8 = 47 \] From recurrence, determine: \[ \alpha + \beta = 1, \quad \alpha \beta = -1 \Rightarrow \text{Required equation with roots } \alpha \text{ and } \frac{1}{\beta} \Rightarrow x^2 + x - 1 = 0 \]