If sum of middle terms is S in the expansion of `(2a-a^2/4)^9` then the value(s) of S is/are

Options

A. `((63)/(32))a^(14)(8+a)`
B. `((63)/(32))a^(13)(8+a)`
C. `((63)/(32))a^(14)(8-a)`
D. `((63)/(32))a^(13)(8-a)`

Answer

To find the sum of the middle terms in the expansion of \((2a - \frac{a^2}{4})^9\), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Terms**:
The expression can be rewritten as \((2a + b)^n\) where \(b = -\frac{a^2}{4}\) and \(n = 9\).

2. **Determine the Number of Terms**:
The number of terms in the expansion is \(n + 1 = 10\). Since \(n\) is odd, there will be two middle terms.

3. **Find the Middle Terms**:
The middle terms are the 5th and 6th terms in the expansion. The general term \(T_k\) in the expansion of \((x + y)^n\) is given by:
\[ T_k = \binom{n}{k-1} x^{n-(k-1)} y^{k-1} \]
For our case, the middle terms are:
- \(T_5\) (5th term)
- \(T_6\) (6th term)

4. **Calculate \(T_5\)**:
\[ T_5 = \binom{9}{4} (2a)^{9-4} \left(-\frac{a^2}{4}\right)^4 \]
Simplifying this:
\[ T_5 = \binom{9}{4} (2a)^5 \left(-\frac{a^2}{4}\right)^4 = \binom{9}{4} (2^5 a^5) \left(-\frac{a^8}{256}\right) \]
\[ = \binom{9}{4} \cdot 32 a^5 \cdot \left(-\frac{a^8}{256}\right) = \binom{9}{4} \cdot \left(-\frac{32}{256}\right) a^{13} \]
\[ = \binom{9}{4} \cdot \left(-\frac{1}{8}\right) a^{13} \]

5. **Calculate \(T_6\)**:
\[ T_6 = \binom{9}{5} (2a)^{9-5} \left(-\frac{a^2}{4}\right)^5 \]
Simplifying this:
\[ T_6 = \binom{9}{5} (2a)^4 \left(-\frac{a^2}{4}\right)^5 = \binom{9}{5} (16 a^4) \left(-\frac{a^{10}}{1024}\right) \]
\[ = \binom{9}{5} \cdot 16 a^4 \cdot \left(-\frac{a^{10}}{1024}\right) = \binom{9}{5} \cdot \left(-\frac{16}{1024}\right) a^{14} \]
\[ = \binom{9}{5} \cdot \left(-\frac{1}{64}\right) a^{14} \]

6. **Sum of the Middle Terms**:
Now, we sum \(T_5\) and \(T_6\):
\[ S = T_5 + T_6 = \binom{9}{4} \cdot \left(-\frac{1}{8}\right) a^{13} + \binom{9}{5} \cdot \left(-\frac{1}{64}\right) a^{14} \]

7. **Calculate Binomial Coefficients**:
\(\binom{9}{4} = 126\) and \(\binom{9}{5} = 126\) (since \(\binom{n}{r} = \binom{n}{n-r}\)):
\[ S = 126 \cdot \left(-\frac{1}{8}\right) a^{13} + 126 \cdot \left(-\frac{1}{64}\right) a^{14} \]
\[ = -\frac{126}{8} a^{13} - \frac{126}{64} a^{14} \]

8. **Final Simplification**:
\[ S = -\frac{63}{4} a^{13} - \frac{63}{64} a^{14} \]

### Final Result:
Thus, the sum of the middle terms \(S\) in the expansion of \((2a - \frac{a^2}{4})^9\) is:
\[ S = -\frac{63}{4} a^{13} - \frac{63}{64} a^{14} \]