Find the `4^(t h)`term in the expansion of `(x-2y)^(12)`.

Explanation

To find the 4th term in the expansion of \((x - 2y)^{12}\), we will use the Binomial Theorem. The Binomial Theorem states that:

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

Here, \(p = x\), \(q = -2y\), and \(n = 12\).

### Step-by-Step Solution:

1. **Identify the term we need**: We are looking for the 4th term in the expansion, which corresponds to \(r = 3\) (since the first term corresponds to \(r = 0\)).

2. **Use the Binomial Formula**: The \(r^{th}\) term in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]
Substituting \(n = 12\), \(p = x\), \(q = -2y\), and \(r = 3\):
\[ T_4 = \binom{12}{3} x^{12-3} (-2y)^3 \]

3. **Calculate the binomial coefficient**:
\[ \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

4. **Calculate the powers**:
- \(x^{12-3} = x^9\)
- \((-2y)^3 = -2^3 y^3 = -8y^3\)

5. **Combine the results**:
\[ T_4 = 220 \cdot x^9 \cdot (-8y^3) = 220 \cdot (-8) \cdot x^9 \cdot y^3 \]
\[ T_4 = -1760 x^9 y^3 \]

### Final Answer:
The 4th term in the expansion of \((x - 2y)^{12}\) is:
\[ -1760 x^9 y^3 \]