If `alpha and beta` are the roots of the quadratic equation `4x ^(2) + 2x -1=0` then the value of `sum _(r =1) ^(oo) (a ^(r ) + beta ^(r ))` is :

Options

A. 2
B. 3
C. 6
D. 0

Answer

To solve the problem, we need to find the value of the infinite series \( S = \sum_{r=1}^{\infty} (\alpha^r + \beta^r) \), where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( 4x^2 + 2x - 1 = 0 \).

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

We can find the roots of the quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = 2 \), and \( c = -1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 4 \cdot (-1) = 4 + 16 = 20 \]

Now substituting into the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{20}}{2 \cdot 4} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4} \]

Thus, the roots are:
\[ \alpha = \frac{-1 + \sqrt{5}}{4}, \quad \beta = \frac{-1 - \sqrt{5}}{4} \]

### Step 2: Calculate the sum \( S = \sum_{r=1}^{\infty} (\alpha^r + \beta^r) \)

We can separate the series:
\[ S = \sum_{r=1}^{\infty} \alpha^r + \sum_{r=1}^{\infty} \beta^r \]

Both series are geometric series. The sum of an infinite geometric series is given by:
\[ \text{Sum} = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

For the series involving \( \alpha \):
- First term \( a = \alpha \)
- Common ratio \( r = \alpha \)

Thus,
\[ \sum_{r=1}^{\infty} \alpha^r = \frac{\alpha}{1 - \alpha} \]

For the series involving \( \beta \):
- First term \( a = \beta \)
- Common ratio \( r = \beta \)

Thus,
\[ \sum_{r=1}^{\infty} \beta^r = \frac{\beta}{1 - \beta} \]

### Step 3: Combine the results

Now, we can combine the two results:
\[ S = \frac{\alpha}{1 - \alpha} + \frac{\beta}{1 - \beta} \]

### Step 4: Find \( 1 - \alpha \) and \( 1 - \beta \)

Calculating \( 1 - \alpha \) and \( 1 - \beta \):
\[ 1 - \alpha = 1 - \frac{-1 + \sqrt{5}}{4} = \frac{4 + 1 - \sqrt{5}}{4} = \frac{5 - \sqrt{5}}{4} \]
\[ 1 - \beta = 1 - \frac{-1 - \sqrt{5}}{4} = \frac{4 + 1 + \sqrt{5}}{4} = \frac{5 + \sqrt{5}}{4} \]

### Step 5: Substitute back into the sum \( S \)

Now substituting these values into \( S \):
\[ S = \frac{\alpha}{1 - \alpha} + \frac{\beta}{1 - \beta} = \frac{\frac{-1 + \sqrt{5}}{4}}{\frac{5 - \sqrt{5}}{4}} + \frac{\frac{-1 - \sqrt{5}}{4}}{\frac{5 + \sqrt{5}}{4}} \]

This simplifies to:
\[ S = \frac{-1 + \sqrt{5}}{5 - \sqrt{5}} + \frac{-1 - \sqrt{5}}{5 + \sqrt{5}} \]

### Step 6: Simplify each fraction

To simplify, we can multiply the numerators and denominators:
\[ S = \frac{(-1 + \sqrt{5})(5 + \sqrt{5}) + (-1 - \sqrt{5})(5 - \sqrt{5})}{(5 - \sqrt{5})(5 + \sqrt{5})} \]

Calculating the denominator:
\[ (5 - \sqrt{5})(5 + \sqrt{5}) = 25 - 5 = 20 \]

Calculating the numerator:
\[ (-1 + \sqrt{5})(5 + \sqrt{5}) = -5 - \sqrt{5} + 5\sqrt{5} + 5 = 4 + 4\sqrt{5} \]
\[ (-1 - \sqrt{5})(5 - \sqrt{5}) = -5 + \sqrt{5} - 5\sqrt{5} + 5 = -4 - 4\sqrt{5} \]

Adding these:
\[ (4 + 4\sqrt{5}) + (-4 - 4\sqrt{5}) = 0 \]

Thus, we find:
\[ S = \frac{0}{20} = 0 \]

### Final Answer
The value of \( S = \sum_{r=1}^{\infty} (\alpha^r + \beta^r) = 0 \).