How many middle terms are there in the expansion of `(2x + 3y)^128`

Answer

To find the number of middle terms in the expansion of \((2x + 3y)^{128}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the exponent**: The exponent in the expression \((2x + 3y)^{128}\) is \(n = 128\).

2. **Determine if the exponent is even or odd**: Since \(128\) is an even number, we can proceed to find the middle terms.

3. **Formula for middle terms**:
- If \(n\) is even, the number of middle terms in the expansion is given by \(1\).
- If \(n\) were odd, the number of middle terms would be \(2\).

4. **Conclusion**: Since \(128\) is even, there is exactly **1 middle term** in the expansion of \((2x + 3y)^{128}\).

### Final Answer:
The number of middle terms in the expansion of \((2x + 3y)^{128}\) is **1**.