Find the middle term is the expansion of : `(2x^2 - (1)/(3x^2))^10`

Answer

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

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

### Step 2: Determine the middle term
Since \(n\) is even, the middle term can be found using the formula:
\[ T_{\frac{n}{2} + 1} = T_{6} \]
where \(n = 10\), so \(\frac{n}{2} + 1 = 5 + 1 = 6\).

### Step 3: Use the general term formula
The general term \(T_r\) in the expansion of \((x + y)^n\) is given by:
\[ T_r = \binom{n}{r-1} x^{n-(r-1)} y^{r-1} \]
For \(T_6\), we have \(r = 6\):
\[ T_6 = \binom{10}{5} (2x^2)^{10-5} \left(-\frac{1}{3x^2}\right)^{5} \]

### Step 4: Calculate the binomial coefficient
Calculating \(\binom{10}{5}\):
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 5: Substitute and simplify
Substituting back into the formula:
\[ T_6 = 252 (2x^2)^{5} \left(-\frac{1}{3x^2}\right)^{5} \]
Calculating each term:
\[ (2x^2)^{5} = 2^5 (x^2)^5 = 32x^{10} \]
\[ \left(-\frac{1}{3x^2}\right)^{5} = -\frac{1^5}{3^5 (x^2)^5} = -\frac{1}{243x^{10}} \]

### Step 6: Combine the terms
Now substituting these into \(T_6\):
\[ T_6 = 252 \cdot 32x^{10} \cdot \left(-\frac{1}{243x^{10}}\right) \]
\[ T_6 = 252 \cdot 32 \cdot \left(-\frac{1}{243}\right) \]
Calculating:
\[ T_6 = -\frac{252 \cdot 32}{243} \]

### Step 7: Simplify the fraction
Calculating \(252 \cdot 32\):
\[ 252 \cdot 32 = 8064 \]
Now simplifying:
\[ T_6 = -\frac{8064}{243} \]

### Final Result
Thus, the middle term in the expansion of \((2x^2 - \frac{1}{3x^2})^{10}\) is:
\[ T_6 = -\frac{8064}{243} \]