The middle term of `[2x-(1)/(3x)]^(10)` is -

Options

A. `"^(10)C_(4)(2^(4))/(3^(4))`
B. `"^(-10)C_(5)(2^(5))/(3^(5))`
C. `"^(10)C_(4)(2^(4))/(3^(5))`
D. `"^(10)C_(5)(2^(5))/(3^(5))`

Answer

To find the middle term of the expression \((2x - \frac{1}{3x})^{10}\), we will follow these steps:

### Step 1: Identify the values of \(a\), \(b\), and \(n\)
In the expression \((a + b)^n\), we have:
- \(a = 2x\)
- \(b = -\frac{1}{3x}\)
- \(n = 10\)

### Step 2: Determine the middle term
Since \(n = 10\) (which is even), the middle term can be found using the formula for the \(\frac{n}{2} + 1\)th term. Therefore, the middle term is the \(\frac{10}{2} + 1 = 6\)th term.

### Step 3: Write the formula for the \(r\)th term
The \(r\)th term in the binomial expansion is given by:
\[ T_r = \binom{n}{r-1} a^{n - (r - 1)} b^{r - 1} \]
For the 6th term, we have \(r = 6\).

### Step 4: Substitute values into the formula
Using \(n = 10\), \(a = 2x\), and \(b = -\frac{1}{3x}\):
\[ T_6 = \binom{10}{6-1} (2x)^{10 - (6 - 1)} \left(-\frac{1}{3x}\right)^{6 - 1} \]
This simplifies to:
\[ T_6 = \binom{10}{5} (2x)^{5} \left(-\frac{1}{3x}\right)^{5} \]

### Step 5: Calculate the binomial coefficient and powers
Now, we calculate:
\[ T_6 = \binom{10}{5} (2x)^5 \left(-\frac{1}{3x}\right)^5 \]
Calculating \(\binom{10}{5}\):
\[ \binom{10}{5} = 252 \]
Now, calculate \((2x)^5\):
\[ (2x)^5 = 32x^5 \]
And \(\left(-\frac{1}{3x}\right)^5\):
\[ \left(-\frac{1}{3x}\right)^5 = -\frac{1}{243x^5} \]

### Step 6: Combine the terms
Now, substituting these values back into the expression for \(T_6\):
\[ T_6 = 252 \cdot 32x^5 \cdot \left(-\frac{1}{243x^5}\right) \]
This simplifies to:
\[ T_6 = 252 \cdot 32 \cdot \left(-\frac{1}{243}\right) \]

### Step 7: Calculate the final value
Calculating \(252 \cdot 32\):
\[ 252 \cdot 32 = 8064 \]
Now, we have:
\[ T_6 = -\frac{8064}{243} \]

### Final Answer
Thus, the middle term of the expansion \((2x - \frac{1}{3x})^{10}\) is:
\[ -\frac{8064}{243} \]