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

Answer

To find the middle terms in the expansions of \( (1 + 2x)^{12} \) and \( \left( 2y - \frac{y^2}{2} \right)^{11} \), we will follow these steps:

### Part I: Finding the Middle Term of \( (1 + 2x)^{12} \)

1. **Identify the Power**: The power of the binomial is \( n = 12 \), which is even.

2. **Determine the Position of the Middle Term**: For an even power, the middle term is given by the formula:
\[ \text{Middle Term} = \left( \frac{n}{2} + 1 \right)^{th} \text{ term} \]
Here, \( n = 12 \), so:
\[ \text{Middle Term} = \left( \frac{12}{2} + 1 \right)^{th} = 7^{th} \text{ term} \]

3. **Use the Binomial Expansion Formula**: The general term in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = 1 \), \( b = 2x \), and \( n = 12 \). For the 7th term, \( r = 6 \) (since \( r \) starts from 0):
\[ T_7 = \binom{12}{6} (1)^{12-6} (2x)^6 \]

4. **Calculate the Middle Term**:
\[ T_7 = \binom{12}{6} (2x)^6 = \binom{12}{6} \cdot 2^6 \cdot x^6 \]
\[ = 924 \cdot 64 \cdot x^6 = 59136x^6 \]

### Part II: Finding the Middle Term of \( \left( 2y - \frac{y^2}{2} \right)^{11} \)

1. **Identify the Power**: The power of the binomial is \( n = 11 \), which is odd.

2. **Determine the Position of the Middle Term**: For an odd power, the middle term is given by the formula:
\[ \text{Middle Term} = \left( \frac{n + 1}{2} \right)^{th} \text{ term} \]
Here, \( n = 11 \), so:
\[ \text{Middle Term} = \left( \frac{11 + 1}{2} \right)^{th} = 6^{th} \text{ term} \]

3. **Use the Binomial Expansion Formula**: The general term in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = 2y \), \( b = -\frac{y^2}{2} \), and \( n = 11 \). For the 6th term, \( r = 5 \):
\[ T_6 = \binom{11}{5} (2y)^{11-5} \left(-\frac{y^2}{2}\right)^5 \]

4. **Calculate the Middle Term**:
\[ T_6 = \binom{11}{5} (2y)^6 \left(-\frac{y^2}{2}\right)^5 \]
\[ = \binom{11}{5} \cdot 2^6 \cdot y^6 \cdot \left(-\frac{y^2}{2}\right)^5 \]
\[ = \binom{11}{5} \cdot 64 \cdot y^6 \cdot \left(-\frac{y^{10}}{32}\right) \]
\[ = \binom{11}{5} \cdot 64 \cdot \left(-\frac{y^{16}}{32}\right) \]
\[ = \binom{11}{5} \cdot (-2) \cdot y^{16} \]
\[ = 462 \cdot (-2) \cdot y^{16} = -924y^{16} \]

### Final Answers:
- The middle term of \( (1 + 2x)^{12} \) is \( 59136x^6 \).
- The middle term of \( \left( 2y - \frac{y^2}{2} \right)^{11} \) is \( -924y^{16} \).