Find middle term in the expansion of `(x - 2y)^(8)` .

Answer

To find the middle term in the expansion of \((x - 2y)^{8}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the values of \(n\)**:
The given expression is \((x - 2y)^{8}\), where \(n = 8\).

2. **Determine the position of the middle term**:
Since \(n\) is even, the middle term can be found using the formula for the middle term in a binomial expansion, which is given by:
\[ \text{Middle Term} = \left(\frac{n}{2} + 1\right)^{\text{th term}} \]
For \(n = 8\):
\[ \text{Middle Term} = \left(\frac{8}{2} + 1\right)^{\text{th term}} = 5^{\text{th term}} \]

3. **Write the general term of the expansion**:
The general term \(T_{r+1}\) in the expansion of \((a + b)^{n}\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^{r} \]
Here, \(a = x\), \(b = -2y\), and \(n = 8\).

4. **Substituting for the 5th term**:
For the 5th term, we have \(r = 4\) (since \(T_{r+1}\) corresponds to \(r\)):
\[ T_{5} = \binom{8}{4} x^{8-4} (-2y)^{4} \]

5. **Calculate \(\binom{8}{4}\)**:
\[ \binom{8}{4} = \frac{8!}{4! \cdot 4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]

6. **Calculate \((-2y)^{4}\)**:
\[ (-2y)^{4} = (-2)^{4} \cdot y^{4} = 16y^{4} \]

7. **Combine the results**:
Now substituting back into the expression for \(T_{5}\):
\[ T_{5} = 70 \cdot x^{4} \cdot 16y^{4} = 1120 x^{4} y^{4} \]

### Final Answer:
The middle term in the expansion of \((x - 2y)^{8}\) is:
\[ 1120 x^{4} y^{4} \]