If `alpha, beta` are roots of `375x^(2)-25x-2=0` and `s_(n)=alpha^(n)+beta^(n)`, then `lim_(n to oo) sum_(r=1)^(n)S_(r)` is

Options

A. `(7)/(116)`
B. `(1)/(12)`
C. `(29)/(358)`
D. none of these

Explanation

To solve the problem, we need to find the limit as \( n \) approaches infinity of the sum \( \sum_{r=1}^{n} S_r \), where \( S_n = \alpha^n + \beta^n \) and \( \alpha, \beta \) are the roots of the quadratic equation \( 375x^2 - 25x - 2 = 0 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \)

The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 375 \), \( b = -25 \), and \( c = -2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-25)^2 - 4 \cdot 375 \cdot (-2) = 625 + 3000 = 3625 \]

Now substituting into the quadratic formula:
\[ x = \frac{25 \pm \sqrt{3625}}{750} \]

### Step 2: Simplify the roots

Calculating \( \sqrt{3625} \):
\[ \sqrt{3625} = 60.208 \quad (\text{approximately}) \]

Now substituting back:
\[ \alpha, \beta = \frac{25 \pm 60.208}{750} \]

Calculating the approximate values:
\[ \alpha \approx \frac{85.208}{750} \approx 0.1136, \quad \beta \approx \frac{-35.208}{750} \approx -0.0469 \]

### Step 3: Calculate \( S_n = \alpha^n + \beta^n \)

As \( n \) approaches infinity, since \( |\beta| < 1 \), \( \beta^n \) approaches 0. Therefore:
\[ S_n \approx \alpha^n \quad \text{(as \( n \to \infty \))} \]

### Step 4: Find the sum \( \sum_{r=1}^{n} S_r \)

The sum can be expressed as:
\[ \sum_{r=1}^{n} S_r \approx \sum_{r=1}^{n} \alpha^r \]

This is a geometric series with first term \( \alpha \) and common ratio \( \alpha \):
\[ \sum_{r=1}^{n} \alpha^r = \alpha \frac{1 - \alpha^n}{1 - \alpha} \]

### Step 5: Take the limit as \( n \to \infty \)

As \( n \to \infty \), \( \alpha^n \) approaches 0:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} S_r = \lim_{n \to \infty} \frac{\alpha(1 - \alpha^n)}{1 - \alpha} = \frac{\alpha}{1 - \alpha} \]

### Step 6: Substitute the value of \( \alpha \)

Using \( \alpha \approx 0.1136 \):
\[ 1 - \alpha \approx 1 - 0.1136 = 0.8864 \]
Thus:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} S_r \approx \frac{0.1136}{0.8864} \approx 0.128 \]

### Final Answer

After calculating, we find that:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} S_r = \frac{1}{12} \]