If `x=1+a+a^(2)+...oo, a lt 1` and `y=1+b+b^(2)+...oo, b lt 1,` then `1+ab+a^(2)b^(2)+...oo :`

Options

A. `(x+y+xy)/(x+y-1)`
B. `(x+y)/(x+y-1)`
C. `(xy)/(x+y-1)`
D. None of these

Explanation

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

### Step-by-step Solution:

1. **Understanding the Series**:
The series \(1 + ab + a^2b^2 + \ldots\) is a geometric series where the first term \(a_1 = 1\) and the common ratio \(r = ab\).

2. **Sum of an Infinite Geometric Series**:
The formula for the sum \(S\) of an infinite geometric series is given by:
\[ S = \frac{a_1}{1 - r} \]
where \(a_1\) is the first term and \(r\) is the common ratio. Here, \(a_1 = 1\) and \(r = ab\).

3. **Applying the Formula**:
Using the formula, we find:
\[ S = \frac{1}{1 - ab} \]

4. **Finding \(ab\)**:
We need to express \(ab\) in terms of \(x\) and \(y\). From the definitions of \(x\) and \(y\):
\[ x = \frac{1}{1 - a} \quad \text{and} \quad y = \frac{1}{1 - b} \]
Rearranging these gives:
\[ a = 1 - \frac{1}{x} \quad \text{and} \quad b = 1 - \frac{1}{y} \]

5. **Calculating \(ab\)**:
Now, we can find \(ab\):
\[ ab = \left(1 - \frac{1}{x}\right)\left(1 - \frac{1}{y}\right) \]
Expanding this:
\[ ab = 1 - \frac{1}{x} - \frac{1}{y} + \frac{1}{xy} \]

6. **Substituting \(ab\) into the Sum**:
Now we substitute \(ab\) back into the sum formula:
\[ S = \frac{1}{1 - \left(1 - \frac{1}{x} - \frac{1}{y} + \frac{1}{xy}\right)} \]
Simplifying the denominator:
\[ S = \frac{1}{\frac{1}{x} + \frac{1}{y} - \frac{1}{xy}} \]
Taking the common denominator:
\[ S = \frac{xy}{y + x - 1} \]

7. **Final Result**:
Thus, the sum of the series \(1 + ab + a^2b^2 + \ldots\) is:
\[ S = \frac{xy}{x + y - 1} \]