If `alpha,beta` are the roots of equation `x^(2)-10x+2=0` and `a_(n)=alpha^(n)-beta^(n)` then `sum_(n=1)^(50)n.((a_(n+1)+2a_(n-1))/(a_(n)))` is more than (a)12500 (b)12250 (c)12750 (d)12000

Options

A. 12500
B. 12250
C. 12750
D. 12000

Explanation

To solve the problem, we need to follow these steps:

1. **Identify the roots of the quadratic equation**: The given equation is \( x^2 - 10x + 2 = 0 \). We can find the roots \( \alpha \) and \( \beta \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -10, c = 2 \).

Substituting these values:
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 8}}{2} = \frac{10 \pm \sqrt{92}}{2} = \frac{10 \pm 2\sqrt{23}}{2} = 5 \pm \sqrt{23} \]
Thus, \( \alpha = 5 + \sqrt{23} \) and \( \beta = 5 - \sqrt{23} \).

2. **Define \( a_n \)**: We have \( a_n = \alpha^n - \beta^n \).

3. **Set up the summation**: We need to evaluate:
\[ \sum_{n=1}^{50} n \cdot \frac{a_{n+1} + 2a_{n-1}}{a_n} \]

4. **Express \( a_{n+1} \) and \( a_{n-1} \)**:
- \( a_{n+1} = \alpha^{n+1} - \beta^{n+1} \)
- \( a_{n-1} = \alpha^{n-1} - \beta^{n-1} \)

5. **Substitute into the summation**:
\[ \sum_{n=1}^{50} n \cdot \frac{(\alpha^{n+1} - \beta^{n+1}) + 2(\alpha^{n-1} - \beta^{n-1})}{\alpha^n - \beta^n} \]

6. **Simplify the expression**:
\[ = \sum_{n=1}^{50} n \cdot \frac{\alpha^{n+1} + 2\alpha^{n-1} - \beta^{n+1} - 2\beta^{n-1}}{\alpha^n - \beta^n} \]
Factor out common terms:
\[ = \sum_{n=1}^{50} n \cdot \frac{\alpha^{n-1}(\alpha + 2) - \beta^{n-1}(\beta + 2)}{\alpha^n - \beta^n} \]

7. **Use the property of roots**: Since \( \alpha \) and \( \beta \) are roots of the equation, we can express \( \alpha^2 + 2 \) and \( \beta^2 + 2 \) in terms of \( \alpha \) and \( \beta \):
- From the equation \( x^2 - 10x + 2 = 0 \), we have:
\[ \alpha^2 = 10\alpha - 2 \quad \text{and} \quad \beta^2 = 10\beta - 2 \]
Thus,
\[ \alpha^2 + 2 = 10\alpha \quad \text{and} \quad \beta^2 + 2 = 10\beta \]

8. **Substitute back into the summation**:
\[ = \sum_{n=1}^{50} n \cdot \frac{10(\alpha^{n-1} - \beta^{n-1})}{\alpha^n - \beta^n} \]

9. **Recognize the pattern**: Notice that \( \frac{a_n}{a_n} = 1 \), leading to:
\[ = 10 \sum_{n=1}^{50} n \]

10. **Calculate the sum of the first 50 natural numbers**:
\[ \sum_{n=1}^{50} n = \frac{n(n+1)}{2} = \frac{50 \cdot 51}{2} = 1275 \]

11. **Final calculation**:
\[ 10 \cdot 1275 = 12750 \]

Thus, the final answer is \( 12750 \), which is more than \( 12750 \).

**Answer**: (c) 12750