Let ` alpha` and ` beta` be the roots of the equation ` 5x^2 + 6x-2 =0`. if ` S_(n ) = alpha ^(n) + beta^(n)`, `n= 1,2,3 ...` then :

Options

A. `5 S _(6) - 6 S_(5)= 2S_(4)`
B. `6S_(6) + 5S_(5) = 2S_(4)`
C. `5S_(6) + 6S_(5) = 2S_(4) `
D. `6S_6 + 5S_5 + 2S_4 =0`

Answer

To solve the problem, we need to find the values of \( S_n = \alpha^n + \beta^n \) where \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( 5x^2 + 6x - 2 = 0 \).

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

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5 \), \( b = 6 \), and \( c = -2 \).

Calculating the discriminant:
\[ D = b^2 - 4ac = 6^2 - 4 \cdot 5 \cdot (-2) = 36 + 40 = 76 \]

Now substituting into the quadratic formula:
\[ \alpha, \beta = \frac{-6 \pm \sqrt{76}}{2 \cdot 5} = \frac{-6 \pm 2\sqrt{19}}{10} = \frac{-3 \pm \sqrt{19}}{5} \]

Thus, the roots are:
\[ \alpha = \frac{-3 + \sqrt{19}}{5}, \quad \beta = \frac{-3 - \sqrt{19}}{5} \]

### Step 2: Use Vieta's Formulas

From Vieta's formulas:
- \( \alpha + \beta = -\frac{b}{a} = -\frac{6}{5} \)
- \( \alpha \beta = \frac{c}{a} = -\frac{2}{5} \)

### Step 3: Establish a recurrence relation for \( S_n \)

Using the recurrence relation:
\[ S_n = \alpha^n + \beta^n \]
We can derive:
\[ S_{n+1} = (\alpha + \beta) S_n - \alpha \beta S_{n-1} \]
Substituting the values from Vieta's:
\[ S_{n+1} = -\frac{6}{5} S_n + \left(-\frac{2}{5}\right) S_{n-1} \]
Multiplying through by 5 to eliminate the fractions:
\[ 5 S_{n+1} = -6 S_n - 2 S_{n-1} \]
Rearranging gives:
\[ 5 S_{n+1} + 6 S_n + 2 S_{n-1} = 0 \]

### Step 4: Find initial conditions

To solve the recurrence, we need initial conditions \( S_0 \) and \( S_1 \):
- \( S_0 = \alpha^0 + \beta^0 = 1 + 1 = 2 \)
- \( S_1 = \alpha + \beta = -\frac{6}{5} \)

### Step 5: Calculate \( S_n \) for \( n = 2, 3, 4, 5, 6 \)

Using the recurrence relation:
1. For \( n = 2 \):
\[ 5 S_2 + 6 S_1 + 2 S_0 = 0 \implies 5 S_2 + 6(-\frac{6}{5}) + 2(2) = 0 \]
\[ 5 S_2 - \frac{36}{5} + 4 = 0 \implies 5 S_2 = \frac{36}{5} - 4 = \frac{36}{5} - \frac{20}{5} = \frac{16}{5} \]
\[ S_2 = \frac{16}{25} \]

2. For \( n = 3 \):
\[ 5 S_3 + 6 S_2 + 2 S_1 = 0 \implies 5 S_3 + 6(\frac{16}{25}) + 2(-\frac{6}{5}) = 0 \]
\[ 5 S_3 + \frac{96}{25} - \frac{12}{5} = 0 \implies 5 S_3 + \frac{96}{25} - \frac{60}{25} = 0 \]
\[ 5 S_3 + \frac{36}{25} = 0 \implies S_3 = -\frac{36}{125} \]

3. For \( n = 4 \):
\[ 5 S_4 + 6 S_3 + 2 S_2 = 0 \implies 5 S_4 + 6(-\frac{36}{125}) + 2(\frac{16}{25}) = 0 \]
\[ 5 S_4 - \frac{216}{125} + \frac{32}{25} = 0 \implies 5 S_4 - \frac{216}{125} + \frac{160}{125} = 0 \]
\[ 5 S_4 - \frac{56}{125} = 0 \implies S_4 = \frac{56}{625} \]

4. For \( n = 5 \):
\[ 5 S_5 + 6 S_4 + 2 S_3 = 0 \implies 5 S_5 + 6(\frac{56}{625}) + 2(-\frac{36}{125}) = 0 \]
\[ 5 S_5 + \frac{336}{625} - \frac{72}{125} = 0 \implies 5 S_5 + \frac{336}{625} - \frac{360}{625} = 0 \]
\[ 5 S_5 - \frac{24}{625} = 0 \implies S_5 = \frac{24}{3125} \]

5. For \( n = 6 \):
\[ 5 S_6 + 6 S_5 + 2 S_4 = 0 \implies 5 S_6 + 6(\frac{24}{3125}) + 2(\frac{56}{625}) = 0 \]
\[ 5 S_6 + \frac{144}{3125} + \frac{112}{625} = 0 \implies 5 S_6 + \frac{144}{3125} + \frac{224}{3125} = 0 \]
\[ 5 S_6 + \frac{368}{3125} = 0 \implies S_6 = -\frac{368}{15625} \]

### Conclusion

The values of \( S_n \) for \( n = 1, 2, 3, 4, 5, 6 \) are:
- \( S_1 = -\frac{6}{5} \)
- \( S_2 = \frac{16}{25} \)
- \( S_3 = -\frac{36}{125} \)
- \( S_4 = \frac{56}{625} \)
- \( S_5 = \frac{24}{3125} \)
- \( S_6 = -\frac{368}{15625} \)