If \(α , β\) are the roots of \(x 2-x-1=0\) and \(A n=α n+β n\), the Arithmetic mean of \(A n-1 \) and \(A n\) is

Let us find the value of \( A n+1 \): \[ α n+1 + β n+1 = (α + β)(α n + β n) - α β (α n-1 + β n-1 ). \] Since \( α + β = 1 \) and \( α β = -1 \), we have: \[ A n+1 = 1 · A n + A n-1 . \] Thus, the recurrence relation becomes: \[ A n + A n-1 2 = A n+1 2 . \]

If \(α , β\) are the roots of \(x... - Solved

Question 34

Mathematics Linear and Quadratic Equations Hard

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

(A) \(2A_{n-1}\)

(B) \(\frac{1}{2} A_{n+1}\)

(C) \(2A_{n-2}\)

(D) None of the above

✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Let us find the value of \( A_{n+1} \): \[ \alpha^{n+1} + \beta^{n+1} = (\alpha + \beta)\left(\alpha^n + \beta^n\right) - \alpha \beta \left(\alpha^{n-1} + \beta^{n-1}\right). \] Since \( \alpha + \beta = 1 \) and \( \alpha \beta = -1 \), we have: \[ A_{n+1} = 1 \cdot A_n + A_{n-1}. \] Thus, the recurrence relation becomes: \[ \frac{A_n + A_{n-1}}{2} = \frac{A_{n+1}}{2}. \]

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Question 34

📘 Step-by-Step Solution & Detailed Explanation:

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