If `(1 + x)^n = C_0+C_1x+C_2x^2 + ... + C_nx^n`,then for n odd,`C_0^2-C_1^2+C_2^2-C_3^2 + ... +(-1) C_n^2`, is equal to

Options

A. 0
B. `2^(2n - 3)`
C. ` ((2n)!)/(2(!n)^(2))`
D. `2^(2n)`

Explanation

To solve the problem, we need to find the value of the expression \( C_0^2 - C_1^2 + C_2^2 - C_3^2 + \ldots + (-1)^n C_n^2 \) given that \( n \) is odd in 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. **Expanding \( (1 - x)^n \)**:
Similarly, the expansion of \( (1 - x)^n \) is:
\[ (1 - x)^n = C_0 - C_1 x + C_2 x^2 - C_3 x^3 + \ldots + (-1)^n C_n x^n \]

3. **Adding the Two Expansions**:
Now, we can add the two expansions:
\[ (1 + x)^n + (1 - x)^n = 2C_0 + 2C_2 x^2 + 2C_4 x^4 + \ldots \]
All odd powers of \( x \) cancel out because they have opposite signs.

4. **Subtracting the Two Expansions**:
Next, we subtract the two expansions:
\[ (1 + x)^n - (1 - x)^n = 2C_1 x + 2C_3 x^3 + 2C_5 x^5 + \ldots \]
Here, all even powers of \( x \) cancel out.

5. **Finding the Coefficient of \( x^n \)**:
To find \( C_0^2 - C_1^2 + C_2^2 - C_3^2 + \ldots + (-1)^n C_n^2 \), we can use the identity:
\[ C_0^2 - C_1^2 + C_2^2 - C_3^2 + \ldots + (-1)^n C_n^2 = \text{Coefficient of } x^n \text{ in } (1+x)^n (1-x)^n \]
This simplifies to:
\[ (1+x)^n (1-x)^n = (1-x^2)^n \]

6. **Evaluating \( (1-x^2)^n \)**:
The coefficient of \( x^n \) in \( (1-x^2)^n \) is 0 when \( n \) is odd because \( x^n \) cannot be formed from \( x^2 \) terms.

7. **Conclusion**:
Therefore, we conclude that:
\[ C_0^2 - C_1^2 + C_2^2 - C_3^2 + \ldots + (-1)^n C_n^2 = 0 \]

### Final Answer:
The value of \( C_0^2 - C_1^2 + C_2^2 - C_3^2 + \ldots + (-1)^n C_n^2 \) for odd \( n \) is **0**.