With usual notations
`C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+…+C_(n-2)C_(n)=`

Options

A. `((2n)!)/((n!)^(2))`
B. `((2n)!)/((n-1)!(n+1)!)`
C. `((2n)!)/((n-2)! (n+2)!)`
D. none

Answer

To solve the problem \( C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n \), we can use a combinatorial identity related to the binomial coefficients.

### Step-by-step Solution:

1. **Understanding the Terms**:
The expression consists of products of binomial coefficients. The term \( C_k \) represents the binomial coefficient \( \binom{n}{k} \).

2. **Rewriting the Expression**:
We can rewrite the expression as:
\[ \sum_{k=0}^{n-2} C_k C_{k+2} \]
This means we are summing the products of pairs of binomial coefficients.

3. **Using the Binomial Coefficient Identity**:
There is a known identity for the sum of products of binomial coefficients:
\[ \sum_{k=0}^{r} C_k C_{r-k} = C_r \]
In our case, we can relate our sum to a specific case of this identity.

4. **Applying the Identity**:
We can express our sum in terms of a shifted index. Notice that:
\[ C_k C_{k+2} = C_k C_{(n-2)-(k-2)} = C_k C_{n-2-k+2} \]
This suggests that we can relate our sum to a binomial coefficient for \( n \).

5. **Finding the Final Expression**:
By applying the identity and adjusting for the indices, we find that:
\[ C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n = C_{n+2} \]
This means that the original expression simplifies to:
\[ \sum_{k=0}^{n-2} C_k C_{k+2} = \frac{1}{2} C_{n+2} \]

6. **Final Result**:
Therefore, the final result of the sum \( C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n \) is:
\[ \frac{1}{2} C_{n+2} \]