`(.^(n)C_(0))^(2)+(.^(n)C_(1))^(2)+(.^(n)C_(2))^(2)+ . . .+(.^(n)C_(n))^(2)` equals

Options

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

Answer

To solve the problem \( \sum_{k=0}^{n} \binom{n}{k}^2 \), we can use the Binomial Theorem and some properties of binomial coefficients. Let's go through the steps:

### Step-by-Step Solution:

1. **Understanding the Sum**: We need to evaluate the sum of the squares of the binomial coefficients:
\[ S = \sum_{k=0}^{n} \binom{n}{k}^2 \]

2. **Using the Binomial Theorem**: Recall the Binomial Theorem states:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
If we set \( x = 1 \) and \( y = 1 \), we get:
\[ (1 + 1)^n = 2^n = \sum_{k=0}^{n} \binom{n}{k} \]

3. **Rearranging the Binomial Theorem**: Now consider the expression \( (1+x)^n(1+x)^n \):
\[ (1+x)^n(1+x)^n = (1+x)^{2n} \]
The left-hand side can be expanded as:
\[ \sum_{k=0}^{n} \binom{n}{k} x^k \sum_{j=0}^{n} \binom{n}{j} x^j = \sum_{m=0}^{2n} \left( \sum_{k=0}^{m} \binom{n}{k} \binom{n}{m-k} \right) x^m \]

4. **Identifying Coefficients**: The coefficient of \( x^m \) in the expansion of \( (1+x)^{2n} \) is \( \binom{2n}{m} \). Therefore, we have:
\[ \sum_{k=0}^{m} \binom{n}{k} \binom{n}{m-k} = \binom{2n}{m} \]

5. **Setting \( m = n \)**: For \( m = n \), we find:
\[ \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k} = \binom{2n}{n} \]
Since \( \binom{n}{n-k} = \binom{n}{k} \), we can rewrite this as:
\[ \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \]

6. **Final Result**: Thus, we conclude that:
\[ S = \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \]

### Conclusion:
The value of \( \sum_{k=0}^{n} \binom{n}{k}^2 \) is \( \binom{2n}{n} \).