If `x=1+a+a^(2)+…….oo,y=1+b+b^(2)…….oo` then `1+ab+a^(2)b^(2)+……….oo,|a|lt1,|b|lt1` is

Options

A. A. `(x+y)/(xy-1)`
B. B. `(x-y)/(xy+1)`
C. C. `(xy)/(x+y-1)`
D. D. `(xy)/(x-y+1)`

Answer

To solve the problem, we need to find the sum of the series \( S = 1 + ab + a^2b^2 + \ldots \) given that \( x = 1 + a + a^2 + \ldots \) and \( y = 1 + b + b^2 + \ldots \) where \( |a| < 1 \) and \( |b| < 1 \).

### Step-by-Step Solution:

1. **Identify the series for \( x \) and \( y \)**:
The series for \( x \) is a geometric series:
\[ x = 1 + a + a^2 + \ldots = \frac{1}{1 - a} \quad \text{(for } |a| < 1\text{)} \]
Similarly, for \( y \):
\[ y = 1 + b + b^2 + \ldots = \frac{1}{1 - b} \quad \text{(for } |b| < 1\text{)} \]

2. **Express \( a \) and \( b \) in terms of \( x \) and \( y \)**:
From the expressions for \( x \) and \( y \), we can derive:
\[ a = 1 - \frac{1}{x} \quad \text{and} \quad b = 1 - \frac{1}{y} \]

3. **Find the sum \( S \)**:
The sum \( S \) can also be expressed as a geometric series:
\[ S = 1 + ab + a^2b^2 + \ldots = \frac{1}{1 - ab} \quad \text{(for } |ab| < 1\text{)} \]

4. **Calculate \( ab \)**:
Substitute the values of \( a \) and \( b \):
\[ ab = \left(1 - \frac{1}{x}\right)\left(1 - \frac{1}{y}\right) = 1 - \frac{1}{x} - \frac{1}{y} + \frac{1}{xy} \]
Rearranging gives:
\[ ab = \frac{xy - x - y + 1}{xy} \]

5. **Substitute \( ab \) into \( S \)**:
Now substitute \( ab \) into the formula for \( S \):
\[ S = \frac{1}{1 - ab} = \frac{1}{1 - \left( \frac{xy - x - y + 1}{xy} \right)} \]
Simplifying this expression:
\[ S = \frac{1}{\frac{xy - (xy - x - y + 1)}{xy}} = \frac{xy}{x + y - 1} \]

6. **Final Result**:
Therefore, the final expression for \( S \) is:
\[ S = \frac{xy}{x + y - 1} \]

### Conclusion:
The answer is \( S = \frac{xy}{x + y - 1} \).