If $\alpha\ and\ \beta$ the roots of $x^{2} - x - 1 = 0\ and\ A_{n} = \alpha^{n} + \beta^{n},\ then\ AM\ of\ A_{n - 1}and\ A_{n}$ is

Question

Dr. Abubacker · Accepted Answer

(b)
Since, $\alpha\ and\ \beta$ the roots of the equation $x^{2} - x - 1 = 0$
$	herefore\ \alpha + \beta = 1\ and\ \alpha\beta = - 1$
Hence, AM of $A_{n - 1}\ and\ A_{n} = \frac{A_{n - 1} + A_{n}}{2}$
$= \frac{\alpha^{n - 1} + \beta^{n - 1} + \alpha^{n} + \beta^{n}}{2}$
$= \frac{\alpha^{n - 1}(1 + \alpha) + \beta^{n - 1}(1 + \beta)}{2}$
$= \frac{\alpha^{n - 1}.\alpha^{2} + \beta^{n - 1}\beta^{2}}{2}$
$= \frac{1}{2}(\alpha^{n + 1} + \beta^{n + 1})$
$= \frac{1}{2}A^{n + 1}$

Q. If $\alpha\ and\ \beta$ the roots of $x^{2} - x - 1 = 0\ and\ A_{n} = \alpha^{n} + \beta^{n},\ then\ AM\ of\ A_{n - 1}and\ A_{n}$ is

Explanation

Reviewed by Dr. Abubacker

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