Let `alpha` and `beta` be the roots of `x^2-6x-2=0` with `alpha>beta` if `a_n=alpha^n-beta^n` for ` n>=1` then the value of `(a_10 - 2a_8)/(2a_9)`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we start by identifying the roots of the quadratic equation \( x^2 - 6x - 2 = 0 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \)
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -6, c = -2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot (-2) = 36 + 8 = 44 \]
Now substituting back into the quadratic formula:
\[ x = \frac{6 \pm \sqrt{44}}{2} = \frac{6 \pm 2\sqrt{11}}{2} = 3 \pm \sqrt{11} \]
Thus, the roots are:
\[ \alpha = 3 + \sqrt{11}, \quad \beta = 3 - \sqrt{11} \]

### Step 2: Define \( a_n \)
Given \( a_n = \alpha^n - \beta^n \).

### Step 3: Establish recurrence relation
From the properties of the roots of the quadratic equation, we can derive a recurrence relation:
\[ a_n = 6a_{n-1} + 2a_{n-2} \]
for \( n \geq 3 \).

### Step 4: Calculate \( a_1 \) and \( a_2 \)
Calculating \( a_1 \) and \( a_2 \):
\[ a_1 = \alpha - \beta = (3 + \sqrt{11}) - (3 - \sqrt{11}) = 2\sqrt{11} \]
\[ a_2 = \alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta) = 2\sqrt{11} \cdot 6 = 12\sqrt{11} \]

### Step 5: Calculate \( a_3 \) and \( a_4 \) using the recurrence relation
Using the recurrence relation:
\[ a_3 = 6a_2 + 2a_1 = 6(12\sqrt{11}) + 2(2\sqrt{11}) = 72\sqrt{11} + 4\sqrt{11} = 76\sqrt{11} \]
\[ a_4 = 6a_3 + 2a_2 = 6(76\sqrt{11}) + 2(12\sqrt{11}) = 456\sqrt{11} + 24\sqrt{11} = 480\sqrt{11} \]

### Step 6: Continue calculating up to \( a_{10} \)
Continuing this process:
\[ a_5 = 6a_4 + 2a_3 = 6(480\sqrt{11}) + 2(76\sqrt{11}) = 2880\sqrt{11} + 152\sqrt{11} = 3032\sqrt{11} \]
\[ a_6 = 6a_5 + 2a_4 = 6(3032\sqrt{11}) + 2(480\sqrt{11}) = 18192\sqrt{11} + 960\sqrt{11} = 19152\sqrt{11} \]
\[ a_7 = 6a_6 + 2a_5 = 6(19152\sqrt{11}) + 2(3032\sqrt{11}) = 114912\sqrt{11} + 6064\sqrt{11} = 120976\sqrt{11} \]
\[ a_8 = 6a_7 + 2a_6 = 6(120976\sqrt{11}) + 2(19152\sqrt{11}) = 725856\sqrt{11} + 38304\sqrt{11} = 764160\sqrt{11} \]
\[ a_9 = 6a_8 + 2a_7 = 6(764160\sqrt{11}) + 2(120976\sqrt{11}) = 4584960\sqrt{11} + 241952\sqrt{11} = 4826912\sqrt{11} \]
\[ a_{10} = 6a_9 + 2a_8 = 6(4826912\sqrt{11}) + 2(764160\sqrt{11}) = 28961472\sqrt{11} + 1528320\sqrt{11} = 30489792\sqrt{11} \]

### Step 7: Calculate the expression \( \frac{a_{10} - 2a_8}{2a_9} \)
Now substituting the values:
\[ \frac{a_{10} - 2a_8}{2a_9} = \frac{(30489792\sqrt{11} - 2 \cdot 764160\sqrt{11})}{2 \cdot 4826912\sqrt{11}} = \frac{(30489792 - 1528320)}{9653824} \]
Calculating the numerator:
\[ 30489792 - 1528320 = 28961472 \]
Thus:
\[ \frac{28961472}{9653824} = 3 \]

### Final Answer
The value of \( \frac{a_{10} - 2a_8}{2a_9} \) is \( 3 \).