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

Options

A. `""^(10)C_(5).2^(5)`
B. `""^(10)C_(6).2^(4)`
C. `""^(10)C_(6).2^(5)(-1)^(5)`
D. None of these

Answer

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

### Step 1: Identify the general term in the expansion
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = 2x \), \( b = -\frac{1}{x} \), and \( n = 10 \).

### Step 2: Write the general term for our specific case
Substituting the values into the formula, we get:
\[ T_{r+1} = \binom{10}{r} (2x)^{10-r} \left(-\frac{1}{x}\right)^r \]

### Step 3: Simplify the general term
Now, we simplify \( T_{r+1} \):
\[ T_{r+1} = \binom{10}{r} (2^{10-r} x^{10-r}) \left(-1^r x^{-r}\right) \]
\[ = \binom{10}{r} 2^{10-r} (-1)^r x^{10-r-r} \]
\[ = \binom{10}{r} 2^{10-r} (-1)^r x^{10-2r} \]

### Step 4: Find the condition for the term to be independent of \( x \)
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 10 - 2r = 0 \]
Solving for \( r \):
\[ 10 = 2r \implies r = 5 \]

### Step 5: Substitute \( r \) back into the general term
Now we substitute \( r = 5 \) into the general term:
\[ T_{6} = \binom{10}{5} 2^{10-5} (-1)^5 \]
\[ = \binom{10}{5} 2^5 (-1)^5 \]

### Step 6: Calculate the values
Calculating \( \binom{10}{5} \) and \( 2^5 \):
\[ \binom{10}{5} = 252, \quad 2^5 = 32 \]
Thus,
\[ T_{6} = 252 \cdot 32 \cdot (-1) = -8064 \]

### Final Answer
The term independent of \( x \) in the expansion of \( (2x - \frac{1}{x})^{10} \) is \( -8064 \).

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