The value of `""^(13)C_(2) +""^(13)C_(3) +""^(13)C_(4) +…+""^(13)C_(13)` is

Options

A. `2^(13)-13`
B. `2^(13)-14`
C. an odd number `ne 2^(13)-12`
D. an even number `ne 2^(13)-14`

Answer

To solve the problem of finding the value of \( \binom{13}{2} + \binom{13}{3} + \binom{13}{4} + \ldots + \binom{13}{13} \), we can use the properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Binomial Theorem**:
The binomial theorem states that:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
For \( x = 1 \) and \( y = 1 \), we have:
\[ (1 + 1)^{13} = \sum_{k=0}^{13} \binom{13}{k} = 2^{13} \]

2. **Calculating Total Sum of Coefficients**:
From the above, we know:
\[ \sum_{k=0}^{13} \binom{13}{k} = 2^{13} = 8192 \]

3. **Separating the Required Terms**:
We need to find:
\[ \sum_{k=2}^{13} \binom{13}{k} = \sum_{k=0}^{13} \binom{13}{k} - \binom{13}{0} - \binom{13}{1} \]

4. **Calculating \( \binom{13}{0} \) and \( \binom{13}{1} \)**:
- \( \binom{13}{0} = 1 \)
- \( \binom{13}{1} = 13 \)

5. **Substituting Values**:
Now substitute these values into the equation:
\[ \sum_{k=2}^{13} \binom{13}{k} = 8192 - 1 - 13 = 8192 - 14 = 8178 \]

6. **Final Result**:
Therefore, the value of \( \binom{13}{2} + \binom{13}{3} + \binom{13}{4} + \ldots + \binom{13}{13} \) is:
\[ \boxed{8178} \]