The value of `""^(14)C_(1) +""^(14)C_(3) +""^(14)C_(5) + …+""^(14)C_(11)` is

Options

A. `2^(14)-1`
B. `2^(14)-14`
C. `2^(12)`
D. `2^(13)-14`

Answer

To solve the problem of finding the value of \( \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \ldots + \binom{14}{11} \), we can use properties of binomial coefficients and the binomial theorem.

### Step-by-Step Solution:

1. **Understanding the Sum of Binomial Coefficients**:
The sum of all binomial coefficients for a given \( n \) is given by:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
For \( n = 14 \):
\[ \sum_{k=0}^{14} \binom{14}{k} = 2^{14} \]

2. **Separating Even and Odd Coefficients**:
The sum of the coefficients for even \( k \) and odd \( k \) can be expressed as:
\[ \sum_{k \text{ even}} \binom{n}{k} + \sum_{k \text{ odd}} \binom{n}{k} = 2^n \]
Since the sum of the even coefficients equals the sum of the odd coefficients, we can denote:
\[ \sum_{k \text{ even}} \binom{14}{k} = \sum_{k \text{ odd}} \binom{14}{k} = \frac{2^{14}}{2} = 2^{13} \]

3. **Finding the Required Sum**:
The sum we need is:
\[ \sum_{k \text{ odd}} \binom{14}{k} = \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \ldots + \binom{14}{11} \]
This sum includes all odd coefficients from \( \binom{14}{1} \) to \( \binom{14}{13} \). We can express it as:
\[ \sum_{k \text{ odd}} \binom{14}{k} = \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \binom{14}{7} + \binom{14}{9} + \binom{14}{11} + \binom{14}{13} \]

4. **Calculating the Last Term**:
We need to find \( \binom{14}{13} \):
\[ \binom{14}{13} = \binom{14}{1} = 14 \]
Thus, we can rewrite the sum of odd coefficients as:
\[ \sum_{k \text{ odd}} \binom{14}{k} = \left( \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \binom{14}{7} + \binom{14}{9} + \binom{14}{11} \right) + 14 \]

5. **Final Calculation**:
Since we know that the total sum of odd coefficients equals \( 2^{13} \):
\[ \sum_{k \text{ odd}} \binom{14}{k} = 8192 \]
Therefore, we can find:
\[ \sum_{k \text{ odd}} \binom{14}{k} - \binom{14}{13} = 8192 - 14 = 8178 \]

### Conclusion:
Thus, the value of \( \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \ldots + \binom{14}{11} \) is:
\[ \boxed{8178} \]