If `alpha`, `beta` are the roots of the equation `ax^(2)+bx+c=0` and `S_(n)=alpha^(n)+beta^(n)`, then `aS_(n+1)+bS_(n)+cS_(n-1)=(n ge 2)`

Options

A. `0`
B. `a+b+c`
C. `(a+b+c)n`
D. `n^(2)abc`

Explanation

To solve the problem, we need to prove that the equation

\[ aS_{n+1} + bS_n + cS_{n-1} = 0 \]

holds for \( n \geq 2 \), where \( S_n = \alpha^n + \beta^n \) and \( \alpha, \beta \) are the roots of the quadratic equation \( ax^2 + bx + c = 0 \).

### Step 1: Express \( S_{n+1} \), \( S_n \), and \( S_{n-1} \)

We know that:

\[ S_n = \alpha^n + \beta^n \]

Thus, we can express \( S_{n+1} \) and \( S_{n-1} \) as follows:

\[ S_{n+1} = \alpha^{n+1} + \beta^{n+1} \]
\[ S_{n-1} = \alpha^{n-1} + \beta^{n-1} \]

### Step 2: Use the properties of roots

From Vieta's formulas, we know:

\[ \alpha + \beta = -\frac{b}{a} \]
\[ \alpha \beta = \frac{c}{a} \]

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

Using the relations for powers of roots, we can derive a recurrence relation:

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

Substituting the values from Vieta's formulas:

\[ S_{n+1} = -\frac{b}{a} S_n - \frac{c}{a} S_{n-1} \]

### Step 4: Multiply the recurrence relation by \( a \)

Now, we multiply the entire equation by \( a \):

\[ aS_{n+1} = -bS_n - cS_{n-1} \]

### Step 5: Rearranging the equation

Rearranging gives us:

\[ aS_{n+1} + bS_n + cS_{n-1} = 0 \]

### Conclusion

Thus, we have shown that:

\[ aS_{n+1} + bS_n + cS_{n-1} = 0 \]

for \( n \geq 2 \).

### Final Answer

The answer is:

\[ \text{Option 1: } 0 \]

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