`(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+….+C_(n)x^(n)` then `C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+…..+C_(n-2)C_(n)` is equal to :

Options

A. `.^(2n)C_(n-1)`
B. `.^(n)C_(n-1)`
C. `.^(2n)C_(n+1)`
D. `.^(n)C_(n+1)`

Answer

To solve the problem, we need to find the value of the expression \( C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n \), where \( C_k = \binom{n}{k} \).

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The binomial expansion of \( (1+x)^n \) is given by:
\[ (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]
where \( C_k = \binom{n}{k} \).

2. **Setting Up the Problem**:
We want to compute the sum \( C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n \). This can be interpreted as the sum of products of coefficients from the binomial expansion.

3. **Using the Binomial Theorem**:
We can express the sum using the identity:
\[ (1+x)^n \cdot (1+x)^n = (1+x)^{2n} \]
This gives us:
\[ (1+x)^n \cdot (1+x)^n = \sum_{k=0}^{2n} C_k' x^k \]
where \( C_k' \) are the new coefficients in the expansion of \( (1+x)^{2n} \).

4. **Identifying the Coefficient**:
The coefficient of \( x^{n-2} \) in \( (1+x)^{2n} \) can be found using the binomial coefficient:
\[ C_{n-2} = \binom{2n}{n-2} \]

5. **Relating the Coefficients**:
The coefficient \( C_{n-2} \) can also be expressed as:
\[ C_{n-2} = C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n \]
This is because the products \( C_r C_{n-r} \) correspond to the combinations of choosing pairs of coefficients from the two expansions.

6. **Final Calculation**:
Thus, we find that:
\[ C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n = \binom{2n}{n-2} \]

### Conclusion:
The final answer is:
\[ \boxed{\binom{2n}{n-2}} \]