Write out the expansions of the following:
(c ) `(x- (y)/(2) )^(4)`

Answer

To expand the expression \((x - \frac{y}{2})^4\) using the Binomial Theorem, we will follow these steps:

### Step 1: Identify the components of the binomial expression
The expression can be rewritten as:
\[ a = x \quad \text{and} \quad b = -\frac{y}{2} \]
with \(n = 4\).

### Step 2: Write the Binomial Expansion Formula
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, we will use:
\[ (x - \frac{y}{2})^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} \left(-\frac{y}{2}\right)^k \]

### Step 3: Calculate each term in the expansion
We will calculate each term for \(k = 0\) to \(k = 4\):

1. **For \(k = 0\)**:
\[ \binom{4}{0} x^{4-0} \left(-\frac{y}{2}\right)^0 = 1 \cdot x^4 \cdot 1 = x^4 \]

2. **For \(k = 1\)**:
\[ \binom{4}{1} x^{4-1} \left(-\frac{y}{2}\right)^1 = 4 \cdot x^3 \cdot \left(-\frac{y}{2}\right) = -2x^3y \]

3. **For \(k = 2\)**:
\[ \binom{4}{2} x^{4-2} \left(-\frac{y}{2}\right)^2 = 6 \cdot x^2 \cdot \left(\frac{y^2}{4}\right) = \frac{3}{2}x^2y^2 \]

4. **For \(k = 3\)**:
\[ \binom{4}{3} x^{4-3} \left(-\frac{y}{2}\right)^3 = 4 \cdot x^1 \cdot \left(-\frac{y^3}{8}\right) = -\frac{1}{2}xy^3 \]

5. **For \(k = 4\)**:
\[ \binom{4}{4} x^{4-4} \left(-\frac{y}{2}\right)^4 = 1 \cdot 1 \cdot \left(\frac{y^4}{16}\right) = \frac{y^4}{16} \]

### Step 4: Combine all the terms
Now we combine all the terms we calculated:
\[ (x - \frac{y}{2})^4 = x^4 - 2x^3y + \frac{3}{2}x^2y^2 - \frac{1}{2}xy^3 + \frac{y^4}{16} \]

### Final Answer
Thus, the expansion of \((x - \frac{y}{2})^4\) is:
\[ x^4 - 2x^3y + \frac{3}{2}x^2y^2 - \frac{1}{2}xy^3 + \frac{y^4}{16} \]

---