The term independent of x in the expansion of
`(1 - x)^(2) (x + (1)/(x))^10`, is

Options

A. `.^(10)C_(5)`
B. `.^(10)C_(6)`
C. `.^(11)C_(5)`
D. `.^(17)C_(7)`

Answer

To find the term independent of \( x \) in the expansion of \( (1 - x)^2 \left( x + \frac{1}{x} \right)^{10} \), we can follow these steps:

### Step 1: Expand \( (1 - x)^2 \)
The expansion of \( (1 - x)^2 \) can be calculated as:
\[ (1 - x)^2 = 1 - 2x + x^2 \]

### Step 2: Expand \( \left( x + \frac{1}{x} \right)^{10} \)
Using the binomial theorem, we can expand \( \left( x + \frac{1}{x} \right)^{10} \):
\[ \left( x + \frac{1}{x} \right)^{10} = \sum_{k=0}^{10} \binom{10}{k} x^{10-k} \left( \frac{1}{x} \right)^k = \sum_{k=0}^{10} \binom{10}{k} x^{10 - 2k} \]

### Step 3: Combine the expansions
Now we need to combine the two expansions:
\[ (1 - 2x + x^2) \left( \sum_{k=0}^{10} \binom{10}{k} x^{10 - 2k} \right) \]

### Step 4: Identify the term independent of \( x \)
We need to find the terms that result in \( x^0 \) (independent of \( x \)). This can happen in the following cases:

1. From \( 1 \) in \( (1 - 2x + x^2) \):
- We need \( 10 - 2k = 0 \) which gives \( k = 5 \).
- The term is \( \binom{10}{5} \).

2. From \( -2x \):
- We need \( 10 - 2k + 1 = 0 \) which gives \( k = 4.5 \) (not possible).

3. From \( x^2 \):
- We need \( 10 - 2k + 2 = 0 \) which gives \( k = 6 \).
- The term is \( \binom{10}{6} \).

### Step 5: Calculate the coefficients
Now we can calculate the coefficients for the valid cases:
- From case 1: \( \binom{10}{5} \)
- From case 3: \( \binom{10}{6} \)

### Step 6: Combine the results
The total coefficient of the term independent of \( x \) is:
\[ \binom{10}{5} + \binom{10}{6} \]

Using the property of binomial coefficients:
\[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \]
we have:
\[ \binom{10}{5} + \binom{10}{6} = \binom{11}{6} \]

### Final Result
Thus, the term independent of \( x \) in the expansion is:
\[ \binom{11}{5} \]