Find the value of `C_0 - [C_1 -2.C_2+ 3.C_3-……..+(-1)^(n-1).n.C_n]`

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we need to find the value of the expression:

\[ C_0 - [C_1 - 2C_2 + 3C_3 - \ldots + (-1)^{n-1} n C_n] \]

where \( C_k \) denotes the binomial coefficient \( \binom{n}{k} \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial coefficients \( C_k \) are given by:
\[ C_k = \binom{n}{k} \]
for \( k = 0, 1, 2, \ldots, n \).

2. **Recall the Binomial Theorem**:
The binomial theorem states that:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]

3. **Differentiate the Binomial Expansion**:
To find the series \( C_1 - 2C_2 + 3C_3 - \ldots + (-1)^{n-1} n C_n \), we differentiate both sides of the binomial expansion with respect to \( x \):
\[ \frac{d}{dx}[(1 + x)^n] = n(1 + x)^{n-1} \]
On the right side, differentiating gives:
\[ C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1} \]

4. **Substituting \( x = -1 \)**:
Now, substituting \( x = -1 \) into the differentiated equation:
\[ n(1 - 1)^{n-1} = 0 \]
This implies:
\[ C_1 - 2C_2 + 3C_3 - \ldots + (-1)^{n-1} n C_n = 0 \]

5. **Substituting into the Original Expression**:
Now, substituting this result back into our original expression:
\[ C_0 - [C_1 - 2C_2 + 3C_3 - \ldots + (-1)^{n-1} n C_n] = C_0 - 0 = C_0 \]

6. **Finding \( C_0 \)**:
The value of \( C_0 \) is \( \binom{n}{0} = 1 \).

### Final Answer:
Thus, the value of the expression is:
\[ \boxed{1} \]