If `alpha, beta` be the roots of `ax^(2)+bx+c=0(a, b, c in R), (c )/(a)lt 1` and `b^(2)-4ac lt 0, f(n)= sum_(r=1)^(n)|alpha|^(r )+|beta|^(r )`, then `lim_(n to oo)f(n)` is equal to

Options

A. `(1)/(sqrt((a)/(c ))-1)`
B. `(1)/(sqrt((a)/(c ))-1)`
C. `(sqrt(c ))/(-sqrt(a)+sqrt(c ))`
D. `(2)/(sqrt((a)/(c ))-1)`

Answer

To solve the problem, we need to find the limit of the function \( f(n) = \sum_{r=1}^{n} |\alpha|^r + |\beta|^r \) as \( n \) approaches infinity, given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( ax^2 + bx + c = 0 \) with certain conditions on \( a, b, \) and \( c \).

### Step 1: Understanding the Roots
Given that \( b^2 - 4ac < 0 \), the roots \( \alpha \) and \( \beta \) are complex and can be expressed as:
\[ \alpha = p + qi \quad \text{and} \quad \beta = p - qi \]
where \( p \) and \( q \) are real numbers.

### Step 2: Finding the Modulus of the Roots
The modulus of the roots can be computed as:
\[ |\alpha| = \sqrt{p^2 + q^2} \quad \text{and} \quad |\beta| = \sqrt{p^2 + q^2} \]
Thus, we have:
\[ |\alpha| = |\beta| = r \quad \text{(say)} \]

### Step 3: Condition on \( c/a \)
From the condition \( \frac{c}{a} < 1 \), we know that:
\[ |\alpha|^2 = \frac{c}{a} < 1 \implies r^2 < 1 \implies r < 1 \]

### Step 4: Summation of the Series
The function \( f(n) \) can be expressed as:
\[ f(n) = \sum_{r=1}^{n} |\alpha|^r + |\beta|^r = \sum_{r=1}^{n} 2|\alpha|^r = 2 \sum_{r=1}^{n} r \]
This is a geometric series where the common ratio is \( r \).

### Step 5: Evaluating the Limit
The sum of a geometric series can be calculated using the formula:
\[ \sum_{r=1}^{n} x^r = x \frac{1 - x^n}{1 - x} \quad \text{for } |x| < 1 \]
Thus, we have:
\[ f(n) = 2 \left( |\alpha| \frac{1 - |\alpha|^n}{1 - |\alpha|} \right) \]
As \( n \to \infty \), \( |\alpha|^n \to 0 \) since \( |\alpha| < 1 \).

### Step 6: Final Limit Calculation
Therefore, we have:
\[ \lim_{n \to \infty} f(n) = 2 \left( \frac{|\alpha|}{1 - |\alpha|} \right) \]
Substituting \( |\alpha|^2 = \frac{c}{a} \):
\[ |\alpha| = \sqrt{\frac{c}{a}} \implies \lim_{n \to \infty} f(n) = \frac{2\sqrt{\frac{c}{a}}}{1 - \sqrt{\frac{c}{a}}} \]

### Conclusion
Thus, the final result is:
\[ \lim_{n \to \infty} f(n) = \frac{2\sqrt{c}}{\sqrt{a}(1 - \sqrt{\frac{c}{a}})} \]