If `x=1-y+y^(2)-y^(3)+` up to infinite terms where `|y| lt1`, then which one of the following is correct ?

Options

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

Answer

To solve the problem, we need to find the sum of the infinite series given by:

\[ x = 1 - y + y^2 - y^3 + \ldots \]

This series is an infinite geometric series where the first term \( a = 1 \) and the common ratio \( r = -y \).

### Step 1: Identify the first term and common ratio
The first term \( a \) is \( 1 \) and the common ratio \( r \) is \( -y \).

### Step 2: Check the condition for convergence
For the infinite geometric series to converge, the absolute value of the common ratio must be less than 1:
\[ |r| < 1 \]
Thus, we have:
\[ |-y| < 1 \]
This simplifies to:
\[ |y| < 1 \]
This condition is given in the problem.

### Step 3: Use the formula for the sum of an infinite geometric series
The sum \( S \) of an infinite geometric series is given by the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values of \( a \) and \( r \):
\[ S = \frac{1}{1 - (-y)} \]
This simplifies to:
\[ S = \frac{1}{1 + y} \]

### Step 4: Set the sum equal to \( x \)
Since we have defined \( x \) as the sum of the series, we can write:
\[ x = \frac{1}{1 + y} \]

### Conclusion
Thus, the correct expression for \( x \) is:
\[ x = \frac{1}{1 + y} \]