Find the 9th term in the expansion of `( 3x - (1)/( 2x ) )^(8), x ne 0`.

Answer

To find the 9th term in the expansion of \( (3x - \frac{1}{2x})^8 \), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case, we can identify:
- \( a = 3x \)
- \( b = -\frac{1}{2x} \)
- \( n = 8 \)

The general term \( T_{k+1} \) in the expansion is given by:

\[ T_{k+1} = \binom{n}{k} a^{n-k} b^k \]

To find the 9th term, we need to set \( k = 8 \) (since the first term corresponds to \( k = 0 \)). Thus, the 9th term \( T_9 \) can be expressed as:

\[ T_9 = \binom{8}{8} (3x)^{8-8} \left(-\frac{1}{2x}\right)^8 \]

Now, we can simplify this step by step:

1. **Calculate the binomial coefficient**:
\[ \binom{8}{8} = 1 \]

2. **Calculate \( (3x)^{8-8} \)**:
\[ (3x)^0 = 1 \]

3. **Calculate \( \left(-\frac{1}{2x}\right)^8 \)**:
\[ \left(-\frac{1}{2x}\right)^8 = \frac{(-1)^8}{(2x)^8} = \frac{1}{256x^8} \]

4. **Combine all parts to find \( T_9 \)**:
\[ T_9 = 1 \cdot 1 \cdot \frac{1}{256x^8} = \frac{1}{256x^8} \]

Thus, the 9th term in the expansion of \( (3x - \frac{1}{2x})^8 \) is:

\[ \frac{1}{256x^8} \]