If `(1+x)^(n) = C_(0) + C_(1)x + C_(2)x^(2) + "….." + C_(n)x^(n)`, then `C_(0) - (C_(0) + C_(1)) +(C_(0) + C_(1) + C_(2)) - (C_(0) + C_(1) + C_(2) + C_(3))+ "….."` `(-1)^(n-1) (C_(0) + C_(1) + "……" + C_(n-1))` is (where n is even integer and `C_(r) = .^(n)C_(r)`)

Options

A. a positive value
B. a negative value
C. divisible by `2^(n-1)`
D. divisible by `2^(n)`

Explanation

To solve the given problem, we start with the binomial expansion of \((1+x)^n\):

\[ (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]

where \(C_r = \binom{n}{r}\).

We need to evaluate the expression:

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

### Step 1: Simplifying the Expression
Let's break down the expression term by term:

1. The first term is \(C_0\).
2. The second term is \(-(C_0 + C_1)\), which simplifies to \(-C_0 - C_1\).
3. The third term is \((C_0 + C_1 + C_2)\), which simplifies to \(C_0 + C_1 + C_2\).
4. The fourth term is \(-(C_0 + C_1 + C_2 + C_3)\), which simplifies to \(-C_0 - C_1 - C_2 - C_3\).

Continuing this pattern, we can see that each pair of terms will contribute a negative value of the next odd indexed coefficient.

### Step 2: Grouping Terms
Grouping the terms gives us:

\[ C_0 - (C_0 + C_1) + (C_0 + C_1 + C_2) - (C_0 + C_1 + C_2 + C_3) + \ldots \]

This can be rearranged to:

\[ -C_1 - C_3 - C_5 - \ldots - C_{n-1} \]

### Step 3: Using Binomial Theorem
Now, we can use the binomial theorem to evaluate the sums:

1. When \(x = 1\):
\[ (1 + 1)^n = 2^n = C_0 + C_1 + C_2 + \ldots + C_n \]

2. When \(x = -1\):
\[ (1 - 1)^n = 0 = C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n \]

### Step 4: Setting Up the Equations
Let’s denote the two equations we derived:

- Equation 1:
\[ C_0 + C_1 + C_2 + \ldots + C_n = 2^n \]

- Equation 2:
\[ C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n = 0 \]

### Step 5: Subtracting the Equations
Now, subtract Equation 2 from Equation 1:

\[ (2^n) - 0 = (C_0 + C_1 + C_2 + \ldots + C_n) - (C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n) \]

This simplifies to:

\[ 2^n = 2(C_1 + C_3 + C_5 + \ldots) \]

### Step 6: Solving for Odd Coefficients
From this, we can find that:

\[ C_1 + C_3 + C_5 + \ldots = \frac{2^n}{2} = 2^{n-1} \]

### Step 7: Final Expression
Thus, the original expression evaluates to:

\[ -C_1 - C_3 - C_5 - \ldots - C_{n-1} = -\left(2^{n-1}\right) = -2^{n-1} \]

### Conclusion
The final answer is:

\[ \boxed{-2^{n-1}} \]