`(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)!)/(n!)^(2)`
B. `((2n)!)/((n-1)!(n+1)!)`
C. `((2n)!)/((n-2)!(n+2)!)`
D. None of these

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 \( C_k \) represents the binomial coefficients from the expansion of \( (1+x)^n \).

### 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. **Rewriting the Expression**:
We need to evaluate:
\[ S = C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n \]

3. **Using the Identity of Binomial Coefficients**:
We can use the identity \( C_k = \binom{n}{k} \) to rewrite \( S \):
\[ S = \sum_{k=0}^{n-2} C_k C_{k+2} = \sum_{k=0}^{n-2} \binom{n}{k} \binom{n}{k+2} \]

4. **Applying Vandermonde's Identity**:
We can express the sum using Vandermonde's identity:
\[ \sum_{k=0}^{r} \binom{m}{k} \binom{n}{r-k} = \binom{m+n}{r} \]
In our case, we can adjust the indices to fit this identity.

5. **Finding the Coefficient**:
To find \( S \), we consider the expansion of \( (1+x)^n \) and \( (1+x)^n \) again, and we multiply them:
\[ (1+x)^n \cdot (1+x)^n = (1+x)^{2n} \]
The coefficient of \( x^{n-2} \) in \( (1+x)^{2n} \) gives us \( C_{n-2}^{2n} \).

6. **Calculating the Coefficient**:
The coefficient of \( x^{n-2} \) in \( (1+x)^{2n} \) is:
\[ C_{n-2} = \binom{2n}{n-2} = \frac{(2n)!}{(n-2)!(n+2)!} \]

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