If `x = 1 + y + y^(2) + y^(3)+…"to"oo`, then y is

Options

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

Answer

To solve the equation \( x = 1 + y + y^2 + y^3 + \ldots \) to infinity, we can recognize that this is an infinite geometric series. Let's break down the steps to find the value of \( y \).

### Step-by-Step Solution:

1. **Identify the Series**:
The series given is \( 1 + y + y^2 + y^3 + \ldots \). This is an infinite geometric series where:
- The first term \( a = 1 \)
- The common ratio \( r = y \)

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

3. **Apply the Formula**:
Substituting the values of \( a \) and \( r \) into the formula, we get:
\[ S = \frac{1}{1 - y} \]
Therefore, we can express \( x \) in terms of \( y \):
\[ x = \frac{1}{1 - y} \]

4. **Rearranging the Equation**:
To find \( y \), we can rearrange the equation:
\[ 1 - y = \frac{1}{x} \]
Now, we can isolate \( y \):
\[ y = 1 - \frac{1}{x} \]

5. **Simplifying the Expression**:
We can simplify this further:
\[ y = \frac{x - 1}{x} \]

### Final Answer:
Thus, the value of \( y \) is:
\[ y = \frac{x - 1}{x} \]