If `(1 + x)^(n) = C_(0) + C_(1)x + C_(2) x^(2) + …+ C_(n) x^(n)`, then for n odd,
`C_(1)^(2) + C_(3)^(2) + C_(5)^(2) +....+ C_(n)^(2)` is equal to

Options

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

Answer

To solve the problem, we need to find the value of the sum \( C_1^2 + C_3^2 + C_5^2 + \ldots + C_n^2 \) where \( n \) is an odd number, and \( C_k \) are the binomial coefficients from the expansion of \( (1 + x)^n \).

### Step-by-Step Solution:

1. **Understanding Binomial Coefficients**:
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. **Identifying Coefficients**:
For odd \( n \), the coefficients can be paired as follows:
\[ C_k = C_{n-k} \]
This means:
- \( C_0 = C_n \)
- \( C_1 = C_{n-1} \)
- \( C_2 = C_{n-2} \)
- \( C_3 = C_{n-3} \)
- and so on.

3. **Setting Up the Equation**:
We can express the sum of the squares of the odd-indexed coefficients:
\[ S = C_1^2 + C_3^2 + C_5^2 + \ldots + C_n^2 \]
By pairing the coefficients, we can also write:
\[ S = C_0^2 + C_2^2 + C_4^2 + \ldots + C_{n-1}^2 \]

4. **Combining the Sums**:
Adding both expressions for \( S \):
\[ 2S = (C_0^2 + C_1^2 + C_2^2 + C_3^2 + \ldots + C_n^2) \]

5. **Finding the Total Sum of Squares**:
We know from the binomial theorem that:
\[ (1 + 1)^n = 2^n = C_0 + C_1 + C_2 + \ldots + C_n \]
To find the sum of the squares, we can use the identity:
\[ (C_0 + C_1 + \ldots + C_n)^2 = C_0^2 + C_1^2 + \ldots + C_n^2 + 2 \sum_{i < j} C_i C_j \]
The sum of products \( \sum_{i < j} C_i C_j \) can be derived from \( (1 + 1)^n \) and is equal to \( 2^{2n-1} \).

6. **Calculating \( S \)**:
Therefore, we can express:
\[ 2^n = C_0 + C_1 + C_2 + \ldots + C_n \]
Squaring gives:
\[ 2^{2n} = C_0^2 + C_1^2 + \ldots + C_n^2 + 2 \cdot 2^{2n-1} \]
Simplifying, we find:
\[ C_0^2 + C_1^2 + \ldots + C_n^2 = 2^{2n} - 2^{2n-1} = 2^{2n-1} \]

7. **Final Calculation**:
Thus, substituting back into our equation for \( S \):
\[ 2S = 2^{2n-1} \implies S = \frac{2^{2n-1}}{2} = 2^{2n-2} \]

### Conclusion:
The value of \( C_1^2 + C_3^2 + C_5^2 + \ldots + C_n^2 \) for odd \( n \) is:
\[ \boxed{2^{2n-2}} \]