If `(1+x)^(n) = C_(0)+C_(1)x + C_(2) x^(2) +...+C_(n)x^(n)`
then ` C_(0)""^(2)+C_(1)""^(2) + C_(2)""^(2) +...+C_(n)""^(2)` is equal to

Options

A. `2^(2n-2)`
B. `2^(n)`
C. `((2n)!)/(2(2!)1^(2))`
D. `((2n)!)/((n!)^(2))`

Explanation

To solve the problem, we need to find the value of \( C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2 \) given that \( (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \).

### Step-by-Step Solution:

1. **Understanding the Coefficients**:
The coefficients \( C_k \) in the expansion of \( (1+x)^n \) are given by the binomial theorem:
\[ C_k = \binom{n}{k} \]
where \( k = 0, 1, 2, \ldots, n \).

2. **Finding \( C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2 \)**:
We need to compute:
\[ C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2 = \sum_{k=0}^{n} \left( \binom{n}{k} \right)^2 \]

3. **Using the Identity for the Sum of Squares of Binomial Coefficients**:
There is a known combinatorial identity that states:
\[ \sum_{k=0}^{n} \left( \binom{n}{k} \right)^2 = \binom{2n}{n} \]
This identity can be derived from the combinatorial interpretation of choosing \( n \) items from \( 2n \) items where we consider the selection of \( n \) items from two groups of \( n \).

4. **Final Result**:
Therefore, we can conclude that:
\[ C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2 = \binom{2n}{n} \]

### Conclusion:
The value of \( C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2 \) is equal to \( \binom{2n}{n} \).