`""^(16)C_(1)-2""^(16)C_(2)+3""^(16)C_(3)…. - 16""^(16)C_(16)=`

Options

A. `2^(15)`
B. `2^(16)-1`
C. 0
D. None

Answer

To solve the expression \( 16C_1 - 2 \cdot 16C_2 + 3 \cdot 16C_3 - \ldots - 16 \cdot 16C_{16} \), we will use the Binomial Theorem and properties of binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression can be rewritten in summation notation as:
\[ \sum_{k=1}^{16} (-1)^{k+1} k \cdot \binom{16}{k} \]
This indicates that we are summing terms where each term is multiplied by the binomial coefficient \( \binom{16}{k} \) and an alternating sign.

2. **Using the Binomial Theorem**:
The Binomial Theorem states that:
\[ (1 - x)^{n} = \sum_{k=0}^{n} \binom{n}{k} (-x)^k \]
Differentiating both sides with respect to \( x \):
\[ -n(1 - x)^{n-1} = \sum_{k=1}^{n} k \cdot \binom{n}{k} (-x)^{k-1} \]

3. **Substituting Values**:
For \( n = 16 \), we have:
\[ -16(1 - x)^{15} = \sum_{k=1}^{16} k \cdot \binom{16}{k} (-x)^{k-1} \]
Now, we will multiply both sides by \( -x \):
\[ 16x(1 - x)^{15} = \sum_{k=1}^{16} k \cdot \binom{16}{k} (-x)^{k} \]

4. **Evaluating at \( x = 1 \)**:
Substitute \( x = 1 \):
\[ 16 \cdot 1 \cdot (1 - 1)^{15} = \sum_{k=1}^{16} k \cdot \binom{16}{k} (-1)^{k} \]
The left-hand side becomes:
\[ 16 \cdot 1 \cdot 0 = 0 \]
Thus, we have:
\[ 0 = 16C_1 - 2 \cdot 16C_2 + 3 \cdot 16C_3 - \ldots - 16 \cdot 16C_{16} \]

5. **Conclusion**:
Therefore, the value of the expression \( 16C_1 - 2 \cdot 16C_2 + 3 \cdot 16C_3 - \ldots - 16 \cdot 16C_{16} \) is:
\[ \boxed{0} \]