If `C_(0) , C_(1) , C_(2) ,…, C_(n) ` are coefficients in the
binomial expansion of `(1 + x)^(n)` and n is even , then
`C_(0)^(2)-C_(1)^(2)+C_(2)^(2)+C_(3)^(2)+...+ (-1)^(n)C_(n)""^(2) ` is equal to .

Options

A. 0
B. `(-1)^(n//2)(n!)/([((n)/(2))!]^(2))`
C. `(-1)^(n) ((2n)!)/((n!)^(2))`
D. `((2n)!)/((n!)^(2))`

Explanation

To solve the problem, we need to evaluate the expression

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

where \(C_k\) are the coefficients in the binomial expansion of \((1 + x)^n\) and \(n\) is even.

### Step-by-Step Solution:

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

2. **Using the Binomial Theorem**:
The binomial theorem states:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
Therefore, we can write:
\[ C_k = \binom{n}{k} \]

3. **Substituting into the Expression**:
We need to evaluate:
\[ \sum_{k=0}^{n} (-1)^k C_k^2 = \sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 \]

4. **Using the Identity for Squared Binomial Coefficients**:
There is a known identity that relates the squared binomial coefficients to another binomial coefficient:
\[ \sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 = (-1)^{n} \binom{n/2}{n/2} \]
when \(n\) is even.

5. **Evaluating the Expression**:
Since \(n\) is even, we can express \(n\) as \(2m\) for some integer \(m\). Thus:
\[ \sum_{k=0}^{2m} (-1)^k \binom{2m}{k}^2 = (-1)^{2m} \binom{m}{m} = 1 \]

6. **Final Result**:
Therefore, the value of the expression is:
\[ C_0^2 - C_1^2 + C_2^2 - C_3^2 + \ldots + (-1)^n C_n^2 = 1 \]

### Conclusion:
The final answer is:
\[ \boxed{1} \]