In the expansion of `(1+x)^(2m)(x/(1-x))^(-2m)` the term independent of x is

Options

A. `"^(2m)C_m`
B. `"^(2m)C_0`
C. `(-1)^m*"^(2m)C_m`
D. none of these

Answer

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

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

### Step 2: Combine the terms
Now we have:
\[ \frac{(1+x)^{2m} (1-x)^{2m}}{x^{2m}} \]

### Step 3: Use the Binomial Theorem
Using the Binomial Theorem, we can expand \( (1+x)^{2m} \) and \( (1-x)^{2m} \):
\[ (1+x)^{2m} = \sum_{r=0}^{2m} \binom{2m}{r} x^r \]
\[ (1-x)^{2m} = \sum_{s=0}^{2m} \binom{2m}{s} (-1)^s x^s \]

### Step 4: Find the product of the two expansions
The product \( (1+x)^{2m} (1-x)^{2m} \) can be calculated as:
\[ \sum_{r=0}^{2m} \sum_{s=0}^{2m} \binom{2m}{r} \binom{2m}{s} (-1)^s x^{r+s} \]

### Step 5: Divide by \( x^{2m} \)
Now we need to divide the entire sum by \( x^{2m} \):
\[ \sum_{r=0}^{2m} \sum_{s=0}^{2m} \binom{2m}{r} \binom{2m}{s} (-1)^s x^{r+s-2m} \]

### Step 6: Find the term independent of \( x \)
To find the term independent of \( x \), we need \( r+s-2m = 0 \) or \( r+s = 2m \). Thus, we need to find pairs \( (r, s) \) such that \( r+s = 2m \).

### Step 7: Calculate the coefficient
The coefficient of the term independent of \( x \) can be calculated as:
\[ \sum_{s=0}^{2m} \binom{2m}{2m-s} \binom{2m}{s} (-1)^s \]
This simplifies to:
\[ \sum_{s=0}^{2m} \binom{2m}{s} \binom{2m}{2m-s} (-1)^s = (-1)^{2m} \cdot \binom{2m}{m} \]
Since \( (-1)^{2m} = 1 \), the coefficient simplifies to:
\[ \binom{2m}{m} \]

### Final Answer
Thus, the term independent of \( x \) in the expansion is:
\[ \binom{2m}{m} \]