Find the 15th term in the expansion of `(2y-(x)/(2))^(18)`

Explanation

To find the 15th term in the expansion of \((2y - \frac{x}{2})^{18}\), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case, we have:
- \(p = 2y\)
- \(q = -\frac{x}{2}\)
- \(n = 18\)

### Step 1: Identify the general term
The general term \(T_{r+1}\) in the expansion can be expressed as:

\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]

Substituting our values, we get:

\[ T_{r+1} = \binom{18}{r} (2y)^{18-r} \left(-\frac{x}{2}\right)^r \]

### Step 2: Find the 15th term
To find the 15th term, we need to set \(r = 14\) (since \(T_{r+1}\) corresponds to \(T_{15}\)):

\[ T_{15} = \binom{18}{14} (2y)^{18-14} \left(-\frac{x}{2}\right)^{14} \]

### Step 3: Simplify the expression
Calculating each component:

1. **Calculate \(\binom{18}{14}\)**:
\[ \binom{18}{14} = \binom{18}{4} = \frac{18!}{4!(18-4)!} = \frac{18 \times 17 \times 16 \times 15}{4 \times 3 \times 2 \times 1} = 3060 \]

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

3. **Calculate \(\left(-\frac{x}{2}\right)^{14}\)**:
\[ \left(-\frac{x}{2}\right)^{14} = (-1)^{14} \left(\frac{x}{2}\right)^{14} = \frac{x^{14}}{2^{14}} = \frac{x^{14}}{16384} \]

### Step 4: Combine all parts
Now, substituting back into the term:

\[ T_{15} = 3060 \cdot 16y^4 \cdot \frac{x^{14}}{16384} \]

### Step 5: Simplify further
Now we simplify:

\[ T_{15} = \frac{3060 \cdot 16}{16384} x^{14} y^4 \]
\[ = \frac{48960}{16384} x^{14} y^4 \]

### Step 6: Final simplification
Now we can simplify \(\frac{48960}{16384}\):

\[ = \frac{765}{256} \]

Thus, the 15th term in the expansion is:

\[ T_{15} = \frac{765}{256} x^{14} y^4 \]

### Final Answer
\[ \text{The 15th term is } \frac{765}{256} x^{14} y^4 \]