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
Step-by-step Solution:
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: \[ \frac{A_n + A_{n-1}}{2} = \frac{A_{n+1}}{2}. \]