The coefficient of `x^(n)`in the expansion of
`((1 + x)/(1-x))^(2), ` is

Options

A. n
B. 2n
C. 3n
D. 4n

Explanation

To find the coefficient of \( x^n \) in the expansion of \( \left( \frac{1+x}{1-x} \right)^2 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \left( \frac{1+x}{1-x} \right)^2 \]
This can be rewritten as:
\[ (1+x)^2 \cdot (1-x)^{-2} \]

### Step 2: Expand \( (1+x)^2 \)
Using the binomial theorem, we expand \( (1+x)^2 \):
\[ (1+x)^2 = 1 + 2x + x^2 \]

### Step 3: Expand \( (1-x)^{-2} \)
Using the binomial series expansion for \( (1-x)^{-n} \), we have:
\[ (1-x)^{-2} = \sum_{k=0}^{\infty} \binom{k+1}{1} x^k = \sum_{k=0}^{\infty} (k+1) x^k \]
This gives us the series:
\[ 1 + 2x + 3x^2 + 4x^3 + \ldots \]

### Step 4: Multiply the expansions
Now we multiply the two expansions:
\[ (1 + 2x + x^2)(1 + 2x + 3x^2 + 4x^3 + \ldots) \]
We need to find the coefficient of \( x^n \) in this product.

### Step 5: Identify the contributions to \( x^n \)
To find the coefficient of \( x^n \), we consider the contributions from:
1. \( 1 \) from \( (1+x)^2 \) and the coefficient of \( x^n \) from \( (1-x)^{-2} \), which is \( n+1 \).
2. \( 2x \) from \( (1+x)^2 \) and the coefficient of \( x^{n-1} \) from \( (1-x)^{-2} \), which is \( n \).
3. \( x^2 \) from \( (1+x)^2 \) and the coefficient of \( x^{n-2} \) from \( (1-x)^{-2} \), which is \( n-1 \).

### Step 6: Combine the contributions
Thus, the coefficient of \( x^n \) is:
\[ (n+1) + 2n + (n-1) = n + 1 + 2n + n - 1 = 4n \]

### Final Answer
The coefficient of \( x^n \) in the expansion of \( \left( \frac{1+x}{1-x} \right)^2 \) is:
\[ \boxed{4n} \]