Find the middle term (s) in the expansion of
(I) `(1+2x)^(12)`
(II) `(2y-(y^(2))/(2))^(11)`

Answer

To find the middle term(s) in the expansions of the given expressions, we will follow a systematic approach for each part.

### Part I: Find the middle term in the expansion of \((1 + 2x)^{12}\)

1. **Identify \(n\)**: In the expression \((1 + 2x)^{12}\), \(n = 12\).

2. **Determine the number of terms**: The number of terms in the expansion is \(n + 1 = 12 + 1 = 13\).

3. **Find the middle term**: Since the number of terms is odd (13), the middle term will be the \(\frac{n}{2} + 1\)th term, which is the \(7\)th term.

4. **Use the binomial formula**: 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, \(a = 1\), \(b = 2x\), and \(n = 12\). For the \(7\)th term, \(r = 6\):
\[ T_7 = \binom{12}{6} (1)^{12-6} (2x)^6 \]

5. **Calculate \(T_7\)**:
\[ T_7 = \binom{12}{6} (2x)^6 = \binom{12}{6} \cdot 2^6 \cdot x^6 \]

6. **Find \(\binom{12}{6}\)**:
\[ \binom{12}{6} = \frac{12!}{6!6!} = 924 \]
Therefore,
\[ T_7 = 924 \cdot 64 \cdot x^6 = 59136x^6 \]

### Part II: Find the middle term in the expansion of \((2y - \frac{y^2}{2})^{11}\)

1. **Identify \(n\)**: In the expression \((2y - \frac{y^2}{2})^{11}\), \(n = 11\).

2. **Determine the number of terms**: The number of terms in the expansion is \(n + 1 = 11 + 1 = 12\).

3. **Find the middle terms**: Since the number of terms is even (12), there will be two middle terms: the \(6\)th and \(7\)th terms.

4. **Calculate the \(6\)th term**:
\[ T_6 = \binom{11}{5} (2y)^{11-5} \left(-\frac{y^2}{2}\right)^5 \]
\[ = \binom{11}{5} (2y)^6 \left(-\frac{y^2}{2}\right)^5 \]

5. **Calculate \(T_6\)**:
\[ = \binom{11}{5} \cdot 2^6 \cdot y^6 \cdot \left(-\frac{1}{2}\right)^5 \cdot y^{10} \]
\[ = \binom{11}{5} \cdot 64 \cdot y^{16} \cdot \left(-\frac{1}{32}\right) \]
\[ = \binom{11}{5} \cdot (-2) \cdot y^{16} \]

6. **Find \(\binom{11}{5}\)**:
\[ \binom{11}{5} = 462 \]
Therefore,
\[ T_6 = 462 \cdot (-2) \cdot y^{16} = -924y^{16} \]

7. **Calculate the \(7\)th term**:
\[ T_7 = \binom{11}{6} (2y)^{11-6} \left(-\frac{y^2}{2}\right)^6 \]
\[ = \binom{11}{6} (2y)^5 \left(-\frac{y^2}{2}\right)^6 \]

8. **Calculate \(T_7\)**:
\[ = \binom{11}{6} \cdot 32 \cdot y^5 \cdot \left(-\frac{1}{64}\right) \cdot y^{12} \]
\[ = \binom{11}{6} \cdot (-\frac{1}{2}) \cdot y^{17} \]

9. **Find \(\binom{11}{6}\)**:
\[ \binom{11}{6} = 462 \]
Therefore,
\[ T_7 = 462 \cdot (-\frac{1}{2}) \cdot y^{17} = -231y^{17} \]

### Final Results:
- The middle term in the expansion of \((1 + 2x)^{12}\) is **\(59136x^6\)**.
- The middle terms in the expansion of \((2y - \frac{y^2}{2})^{11}\) are **\(-924y^{16}\)** and **\(-231y^{17}\)**.