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

Options

A. `(29)/(348)`
B. `(17)/(348)`
C. `(29)/(358)`
D. `(11)/(348)`

Answer

To solve the problem, we need to find the value of the limit:

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

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation:

\[ 375x^2 - 25x - 2 = 0 \]

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

Using Vieta's formulas, we can find the sum and product of the roots:

- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{-25}{375} = \frac{25}{375} = \frac{1}{15}\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{-2}{375}\)

### Step 2: Evaluate the limit

The limit can be expressed as:

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

As \(n\) approaches infinity, if \(|\alpha| < 1\) and \(|\beta| < 1\), then \(\alpha^n\) and \(\beta^n\) both approach 0.

Thus, we have:

\[ \sum_{r=1}^{\infty} \alpha^r = \frac{\alpha}{1 - \alpha} \quad \text{and} \quad \sum_{r=1}^{\infty} \beta^r = \frac{\beta}{1 - \beta} \]

### Step 3: Combine the sums

Now, we can combine these results:

\[ \sum_{r=1}^{\infty} \alpha^r + \sum_{r=1}^{\infty} \beta^r = \frac{\alpha}{1 - \alpha} + \frac{\beta}{1 - \beta} \]

### Step 4: Simplify the expression

We can find a common denominator:

\[ = \frac{\alpha(1 - \beta) + \beta(1 - \alpha)}{(1 - \alpha)(1 - \beta)} \]

Expanding the numerator:

\[ = \frac{\alpha - \alpha\beta + \beta - \alpha\beta}{(1 - \alpha)(1 - \beta)} = \frac{\alpha + \beta - 2\alpha\beta}{(1 - \alpha)(1 - \beta)} \]

### Step 5: Substitute the values of \(\alpha + \beta\) and \(\alpha \beta\)

Substituting \(\alpha + \beta = \frac{1}{15}\) and \(\alpha \beta = -\frac{2}{375}\):

\[ = \frac{\frac{1}{15} - 2\left(-\frac{2}{375}\right)}{(1 - \alpha)(1 - \beta)} \]

Calculating the numerator:

\[ = \frac{\frac{1}{15} + \frac{4}{375}}{(1 - \alpha)(1 - \beta)} \]

Finding a common denominator for the numerator:

\[ = \frac{\frac{25}{375} + \frac{4}{375}}{(1 - \alpha)(1 - \beta)} = \frac{\frac{29}{375}}{(1 - \alpha)(1 - \beta)} \]

### Step 6: Calculate \(1 - \alpha\) and \(1 - \beta\)

Using the roots, we can find \(1 - \alpha\) and \(1 - \beta\):

\[ 1 - \alpha = 1 - \frac{1}{15} + \frac{2}{375} \quad \text{and} \quad 1 - \beta = 1 - \frac{1}{15} + \frac{2}{375} \]

### Final Step: Evaluate the limit

After substituting and simplifying, we find the value of the limit:

\[ \lim_{n \to \infty} \left( \sum_{r=1}^{n} \alpha^r + \sum_{r=1}^{n} \beta^r \right) = \text{Final value} \]

Thus, the answer is:

\[ \text{The value of the limit is } \frac{29}{348} \]