Question

If \( \alpha, \beta \) are the roots of \( x^2 - x - 1 = 0 \), and \( A_n = \alpha^n + \beta^n \), then Arithmetic mean of \( A_{n-1} \) and \( A_n \) is

• A. \( 2A_{n-1} \) (Typo in option)
• B. \( \frac{A_{n+1}}{2} \)
• C. \( 2A_{n-2} \) (Typo in option)
• D. None of the above

Hint:

When given a recurrence relation defined by the roots of a quadratic equation \( ax^2+bx+c=0 \), you can find a linear recurrence relation for \( A_n = \alpha^n + \beta^n \). Multiply the quadratic by \( x^{n-2} \) to get \( ax^n + bx^{n-1} + cx^{n-2} = 0 \). Since both roots satisfy this, so does their sum, giving the recurrence \( aA_n + bA_{n-1} + cA_{n-2} = 0 \). Here, \( x^2-x-1=0 \) gives \( A_{n} - A_{n-1} - A_{n-2} = 0 \), or \( A_n = A_{n-1}+A_{n-2} \).

Answer 1

The Correct Option is B

Since \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - x - 1 = 0 \), they satisfy the equation: \[ \alpha^2 - \alpha - 1 = 0 \implies \alpha^2 = \alpha + 1 \] \[ \beta^2 - \beta - 1 = 0 \implies \beta^2 = \beta + 1 \] We are given \( A_n = \alpha^n + \beta^n \). We need to find the arithmetic mean of \( A_{n-1} \) and \( A_n \), which is \( \frac{A_{n-1} + A_n}{2} \). Let's first find a recurrence relation for \( A_n \). Consider \( A_{n+1} \): \[ A_{n+1} = \alpha^{n+1} + \beta^{n+1} \] \[ = \alpha^{n-1} \cdot \alpha^2 + \beta^{n-1} \cdot \beta^2 \] Substitute \( \alpha^2 = \alpha + 1 \) and \( \beta^2 = \beta + 1 \): \[ A_{n+1} = \alpha^{n-1}(\alpha + 1) + \beta^{n-1}(\beta + 1) \] \[ = (\alpha^n + \alpha^{n-1}) + (\beta^n + \beta^{n-1}) \] \[ = (\alpha^n + \beta^n) + (\alpha^{n-1} + \beta^{n-1}) \] \[ A_{n+1} = A_n + A_{n-1} \] This is the recurrence relation for the sequence (which is related to Lucas numbers). The question asks for the arithmetic mean of \( A_{n-1} \) and \( A_n \), which is \( \frac{A_{n-1} + A_n}{2} \). Using our recurrence relation, we see that \( A_{n-1} + A_n = A_{n+1} \). Therefore, the arithmetic mean is \( \frac{A_{n+1}}{2} \).