Find the fifth expansion of `(2x^(2)-1/(3x^(2)))^(10)`

Answer

To find the fifth term in the expansion of \((2x^2 - \frac{1}{3x^2})^{10}\), we will use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

Where \(\binom{n}{r}\) is the binomial coefficient, which can be calculated as \(\frac{n!}{r!(n-r)!}\).

### Step-by-step Solution:

1. **Identify \(a\), \(b\), and \(n\)**:
- Here, \(a = 2x^2\), \(b = -\frac{1}{3x^2}\), and \(n = 10\).

2. **Find the general term**:
- The general term \(T_{r+1}\) in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
- For our case:
\[ T_{r+1} = \binom{10}{r} (2x^2)^{10-r} \left(-\frac{1}{3x^2}\right)^r \]

3. **Find the fifth term**:
- The fifth term corresponds to \(r = 4\) (since \(T_{r+1}\) corresponds to \(T_5\)).
\[ T_5 = \binom{10}{4} (2x^2)^{10-4} \left(-\frac{1}{3x^2}\right)^4 \]

4. **Calculate \(\binom{10}{4}\)**:
\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]

5. **Calculate \((2x^2)^{6}\)**:
\[ (2x^2)^{6} = 2^6 (x^2)^6 = 64x^{12} \]

6. **Calculate \(\left(-\frac{1}{3x^2}\right)^4\)**:
\[ \left(-\frac{1}{3x^2}\right)^4 = \frac{1}{81x^8} \]

7. **Combine the terms**:
- Now substituting back into \(T_5\):
\[ T_5 = 210 \cdot 64x^{12} \cdot \frac{1}{81x^8} \]
- Simplifying this:
\[ T_5 = 210 \cdot 64 \cdot \frac{x^{12}}{81x^8} = 210 \cdot 64 \cdot \frac{1}{81} x^{4} \]

8. **Calculate the coefficient**:
- Calculate \(210 \cdot 64\):
\[ 210 \cdot 64 = 13440 \]
- Now divide by \(81\):
\[ \frac{13440}{81} \]

9. **Final Result**:
- Thus, the fifth term in the expansion is:
\[ T_5 = \frac{13440}{81} x^4 \]