Let `alpha` and `beta` are two positive roots of `x^(2)-2ax+ab=0` where `0ltblta`, then the value of `S_(n)=1+2((b)/(a))+3((b)/(a))^(2)+……+(n)((b)/(a))^(n-1), AA n in N`
cannot exceed

Options

A. `(alpha)/(beta)`
B. `|(alpha+beta)/(alpha-beta)|`
C. `|(beta)/(alpha)|`
D. `((alpha+beta)/(alpha-beta))^(4)`

Answer

To solve the problem, we will break it down step by step.

### Step 1: Understanding the Roots
We are given the quadratic equation:
\[ x^2 - 2ax + ab = 0 \]
The roots of this equation are denoted as \( \alpha \) and \( \beta \). According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta = 2a \)
- The product of the roots \( \alpha \beta = ab \)

### Step 2: Analyzing the Series
We need to evaluate the series:
\[ S_n = 1 + 2\left(\frac{b}{a}\right) + 3\left(\frac{b}{a}\right)^2 + \ldots + n\left(\frac{b}{a}\right)^{n-1} \]
This series can be recognized as an arithmetic-geometric series.

### Step 3: Formula for the Series
The sum of the first \( n \) terms of an arithmetic-geometric series can be calculated using the formula:
\[ S_n = \frac{a}{(1 - r)} + \frac{d \cdot r}{(1 - r)^2} \]
where \( a \) is the first term, \( d \) is the common difference, and \( r \) is the common ratio.

In our case:
- \( a = 1 \)
- \( d = 1 \) (the difference between consecutive terms)
- \( r = \frac{b}{a} \)

### Step 4: Applying the Formula
Substituting the values into the formula:
\[ S_n = \frac{1}{1 - \frac{b}{a}} + \frac{1 \cdot \frac{b}{a}}{(1 - \frac{b}{a})^2} \]

### Step 5: Simplifying the Expression
We can simplify \( S_n \):
1. The first term becomes:
\[ \frac{1}{1 - \frac{b}{a}} = \frac{a}{a - b} \]

2. The second term becomes:
\[ \frac{\frac{b}{a}}{(1 - \frac{b}{a})^2} = \frac{b/a}{(1 - b/a)^2} = \frac{b}{a} \cdot \frac{a^2}{(a - b)^2} = \frac{ab}{(a - b)^2} \]

Combining these:
\[ S_n = \frac{a}{a - b} + \frac{ab}{(a - b)^2} \]

### Step 6: Finding a Common Denominator
The common denominator is \( (a - b)^2 \):
\[ S_n = \frac{a(a - b) + ab}{(a - b)^2} = \frac{a^2 - ab + ab}{(a - b)^2} = \frac{a^2}{(a - b)^2} \]

### Step 7: Conclusion
Thus, the value of \( S_n \) cannot exceed:
\[ S_n < \frac{a^2}{(a - b)^2} \]