`C_0C_2 + C_1C_3 +C_2C_4+……..+C_(n-2) C_n = `

Options

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

Answer

To solve the problem \( C_0C_2 + C_1C_3 + C_2C_4 + \ldots + C_{n-2}C_n \), we will use the properties of binomial coefficients and the binomial theorem.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the sum of products of binomial coefficients in the form \( C_k C_{n-k} \) for \( k = 0, 1, 2, \ldots, n-2 \).

2. **Using Binomial Theorem**:
Recall the binomial expansion:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]
This can be written as:
\[ (1 + x)^n = \sum_{k=0}^{n} C_k x^k \]

3. **Constructing a Second Expansion**:
We can also consider the expansion:
\[ (1 + x^2)^n = C_0 + C_1 x^2 + C_2 x^4 + \ldots + C_n x^{2n} \]
This gives us:
\[ (1 + x^2)^n = \sum_{k=0}^{n} C_k x^{2k} \]

4. **Multiplying the Two Expansions**:
Now, we multiply the two expansions:
\[ (1 + x)^n (1 + x^2)^n \]
This results in:
\[ (1 + x)^{n} (1 + x^2)^{n} = (1 + x + x^2 + x^3 + \ldots)^n \]
The product will yield terms that correspond to the coefficients \( C_k C_{n-k} \).

5. **Finding the Coefficient of \( x^n \)**:
The left-hand side will give us a series of coefficients that include \( C_k C_{n-k} \). The coefficient of \( x^n \) in this product will be:
\[ \sum_{k=0}^{n} C_k C_{n-k} \]

6. **Using the Hockey Stick Identity**:
The sum \( C_k C_{n-k} \) can be simplified using the Hockey Stick identity, which states that:
\[ C_k C_{n-k} = C_{n+1} \text{ (for appropriate values)} \]

7. **Final Result**:
After simplifying, we find that:
\[ C_0C_2 + C_1C_3 + C_2C_4 + \ldots + C_{n-2}C_n = 2nC_{n-2} \]

### Conclusion:
Thus, the final answer is:
\[ C_0C_2 + C_1C_3 + C_2C_4 + \ldots + C_{n-2}C_n = 2nC_{n-2} \]