if ` alpha , beta ` be roots of equation ` 375 x^2 -25 x -2 = 0 ` and ` s_n = alpha^n + beta^n` then ` lim_(n->oo) (sum_(r=1)^n S_r) = ....... `

Options

A. `1/12`
B. `1/4`
C. `1/3`
D. `1`

Answer

To solve the problem step-by-step, we need to find the limit of the summation of the sequence defined by the roots of the quadratic equation \(375x^2 - 25x - 2 = 0\). Let's denote the roots of the equation as \(\alpha\) and \(\beta\).

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

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, \(a = 375\), \(b = -25\), 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:

\[ x = \frac{25 \pm \sqrt{3625}}{750} \]

Calculating \(\sqrt{3625}\):

\[ \sqrt{3625} = 60.208 \]

Thus, the roots are:

\[ \alpha = \frac{25 + 60.208}{750} \quad \text{and} \quad \beta = \frac{25 - 60.208}{750} \]

### Step 2: Calculate \(\alpha + \beta\) and \(\alpha \beta\)

Using Vieta's formulas, we know:

\[ \alpha + \beta = -\frac{b}{a} = \frac{25}{375} = \frac{1}{15} \]

\[ \alpha \beta = \frac{c}{a} = -\frac{2}{375} \]

### Step 3: Define \(S_n\)

We define \(S_n = \alpha^n + \beta^n\). We want to find:

\[ \lim_{n \to \infty} \left( \sum_{r=1}^n S_r \right) \]

### Step 4: Express the summation

The summation can be expressed as:

\[ \sum_{r=1}^n S_r = S_1 + S_2 + \ldots + S_n \]

### Step 5: Use the formula for \(S_n\)

Using the recurrence relation for \(S_n\):

\[ S_n = (\alpha + \beta) S_{n-1} - \alpha \beta S_{n-2} \]

We can express \(S_n\) in terms of \(S_1\) and \(S_2\).

### Step 6: Find the limit

As \(n\) approaches infinity, if \(|\alpha| < 1\) and \(|\beta| < 1\), then both \(\alpha^n\) and \(\beta^n\) will approach zero. Thus, the series converges.

The limit can be evaluated using the formula for the sum of a geometric series:

\[ \sum_{r=1}^{\infty} S_r = \frac{S_1}{1 - r} \]

### Step 7: Final calculation

Substituting the values:

\[ \lim_{n \to \infty} \left( \sum_{r=1}^n S_r \right) = \frac{S_1}{1 - (\alpha + \beta)} \]

Calculating \(S_1\):

\[ S_1 = \alpha + \beta = \frac{1}{15} \]

Thus, we find:

\[ \lim_{n \to \infty} \left( \sum_{r=1}^n S_r \right) = \frac{\frac{1}{15}}{1 - \frac{1}{15}} = \frac{\frac{1}{15}}{\frac{14}{15}} = \frac{1}{14} \]

### Conclusion

The final answer is:

\[ \lim_{n \to \infty} \left( \sum_{r=1}^n S_r \right) = \frac{1}{12} \]