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 need to find the value of \( S_n = 1 + 2\left(\frac{b}{a}\right) + 3\left(\frac{b}{a}\right)^2 + \ldots + n\left(\frac{b}{a}\right)^{n-1} \) and determine the upper limit for \( S_n \).

### Step 1: Identify the series
The series can be recognized as a weighted geometric series where the first term is \( 1 \) and the common ratio is \( r = \frac{b}{a} \).

### Step 2: Use the formula for the sum of a weighted geometric series
The sum of the series can be expressed using the formula for the sum of a weighted geometric series:
\[ S_n = \sum_{k=1}^{n} k r^{k-1} = \frac{1 - r^n}{(1 - r)^2} - \frac{n r^n}{1 - r} \]
where \( r = \frac{b}{a} \).

### Step 3: Substitute \( r \) into the formula
Substituting \( r = \frac{b}{a} \) into the formula gives:
\[ S_n = \frac{1 - \left(\frac{b}{a}\right)^n}{\left(1 - \frac{b}{a}\right)^2} - \frac{n \left(\frac{b}{a}\right)^n}{1 - \frac{b}{a}} \]

### Step 4: Simplify the expression
Now, we simplify the expression:
\[ S_n = \frac{1 - \left(\frac{b}{a}\right)^n}{\left(\frac{a - b}{a}\right)^2} - \frac{n \left(\frac{b}{a}\right)^n}{\frac{a - b}{a}} \]
This simplifies to:
\[ S_n = \frac{a^2(1 - \left(\frac{b}{a}\right)^n)}{(a - b)^2} - \frac{n b^n}{a^{n-1}(a - b)} \]

### Step 5: Find the limit as \( n \to \infty \)
As \( n \to \infty \), if \( \left|\frac{b}{a}\right| < 1 \), the term \( \left(\frac{b}{a}\right)^n \) approaches \( 0 \). Therefore:
\[ S_n \to \frac{a^2}{(a - b)^2} \]

### Step 6: Determine the maximum value
Thus, the maximum value that \( S_n \) can reach is:
\[ S_n \leq \frac{a^2}{(a - b)^2} \]

### Conclusion
The value of \( S_n \) cannot exceed \( \frac{a^2}{(a - b)^2} \).