If `x = 1 + a + a^(2) + a^(3) +…"to" oo (|a| lt 1) and y = 1 b + b^(2) + b^(3) + …"to" oo (|b| lt 1)`, then `1 + ab + a^(2)b^(2) + a^(3)b^(3) + … oo` is equal to

Options

A. `(xy)/(x + y - 1)`
B. `(x + y)/(x + y + 1)`
C. `(x - y)/(x - y + 1)`
D. none

Answer

To solve the given problem, we need to find the value of the infinite series \( S = 1 + ab + a^2b^2 + a^3b^3 + \ldots \) under the conditions that \( |a| < 1 \) and \( |b| < 1 \).

### Step-by-step Solution:

1. **Identify the series**: The series \( S = 1 + ab + a^2b^2 + a^3b^3 + \ldots \) is a geometric series where the first term \( a_1 = 1 \) and the common ratio \( r = ab \).

2. **Sum of a geometric series**: The sum of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a_1}{1 - r} \]
where \( |r| < 1 \).

3. **Apply the formula**: Here, \( a_1 = 1 \) and \( r = ab \). Therefore, we have:
\[ S = \frac{1}{1 - ab} \]

4. **Final expression**: Thus, the value of the series \( S \) is:
\[ S = \frac{1}{1 - ab} \]

### Conclusion:
The value of \( 1 + ab + a^2b^2 + a^3b^3 + \ldots \) is equal to \( \frac{1}{1 - ab} \).