If `alpha, beta` are the roots of `6x^(2)-2x+1=0 and s_(n) =alpha^(n)+beta^(n)` then `underset(n rarr oo)L underset (r=1)overset(n)Sigma s_(r)` is

Options

A. `(5)/(17)`
B. 0
C. `(3)/(37)`
D. none

Answer

To solve the problem, we need to find the limit as \( n \) approaches infinity of the summation from \( r = 1 \) to \( n \) of \( s_r \), where \( s_n = \alpha^n + \beta^n \) and \( \alpha, \beta \) are the roots of the quadratic equation \( 6x^2 - 2x + 1 = 0 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \) of the quadratic equation
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \( 6x^2 - 2x + 1 = 0 \):
- \( a = 6 \)
- \( b = -2 \)
- \( c = 1 \)

Calculating the discriminant:
\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 6 \cdot 1 = 4 - 24 = -20 \]
Since the discriminant is negative, the roots are complex:
\[ \alpha, \beta = \frac{2 \pm \sqrt{-20}}{12} = \frac{2 \pm 2i\sqrt{5}}{12} = \frac{1 \pm i\sqrt{5}}{6} \]

### Step 2: Calculate \( \alpha + \beta \) and \( \alpha \beta \)
Using Vieta's formulas:
- \( \alpha + \beta = -\frac{b}{a} = \frac{2}{6} = \frac{1}{3} \)
- \( \alpha \beta = \frac{c}{a} = \frac{1}{6} \)

### Step 3: Define \( s_n \)
We have:
\[ s_n = \alpha^n + \beta^n \]

### Step 4: Find the limit of the summation
We need to evaluate:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} s_r \]
This can be rewritten using the formula for \( s_n \):
\[ \sum_{r=1}^{n} s_r = s_1 + s_2 + \ldots + s_n \]

### Step 5: Find \( s_1 \) and \( s_2 \)
Calculating \( s_1 \) and \( s_2 \):
- \( s_1 = \alpha + \beta = \frac{1}{3} \)
- \( s_2 = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{1}{3}\right)^2 - 2 \cdot \frac{1}{6} = \frac{1}{9} - \frac{1}{3} = \frac{1}{9} - \frac{3}{9} = -\frac{2}{9} \)

### Step 6: Use the recurrence relation for \( s_n \)
Using the recurrence relation:
\[ s_n = \left(\frac{1}{3}\right)s_{n-1} - \left(\frac{1}{6}\right)s_{n-2} \]
We can express \( s_n \) in terms of previous terms.

### Step 7: Analyze the behavior as \( n \to \infty \)
As \( n \) approaches infinity, the terms \( s_n \) will tend to zero because the roots \( \alpha \) and \( \beta \) are less than 1 in magnitude (since they are complex). Therefore, the summation will converge.

### Step 8: Final limit calculation
Using the properties of converging series:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} s_r \to 0 \]

Thus, the final answer is:
\[ \boxed{0} \]