The value of `sum_(r=0)^(2n)(-1)^(r)*(""^(2n)C_(r))^(2)` is equal to :

Options

A. `""^(4n)C_(2n)`
B. `""^(2n)C_(n)`
C. `(-1)^(n)*""^(2n)C_(n)`
D. None of these

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{r=0}^{2n} (-1)^r \binom{2n}{r}^2 \]

### Step 1: Understanding the Binomial Coefficient
The binomial coefficient \(\binom{2n}{r}\) represents the number of ways to choose \(r\) elements from a set of \(2n\) elements. The square of this coefficient, \(\binom{2n}{r}^2\), can be interpreted combinatorially as the number of ways to choose \(r\) elements from two independent sets of \(2n\) elements.

### Step 2: Applying the Binomial Theorem
We can use the binomial theorem, which states that:

\[ (1 + x)^{2n} = \sum_{r=0}^{2n} \binom{2n}{r} x^r \]

### Step 3: Evaluating the Alternating Sum
To evaluate the alternating sum, we can substitute \(x = -1\) into the binomial expansion:

\[ (1 - 1)^{2n} = 0 = \sum_{r=0}^{2n} \binom{2n}{r} (-1)^r \]

This shows that the sum of the binomial coefficients with alternating signs is zero.

### Step 4: Squaring the Binomial Coefficients
Now we need to consider the square of the binomial coefficients. We can express the sum \(S\) in terms of generating functions. The generating function for \(\binom{2n}{r}\) is:

\[ (1 + x)^{2n} (1 - x)^{2n} = (1 - x^2)^{2n} \]

### Step 5: Coefficient Extraction
The coefficient of \(x^{2n}\) in the expansion of \((1 - x^2)^{2n}\) gives us the required sum. The expansion can be written as:

\[ (1 - x^2)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k x^{2k} \]

To find the coefficient of \(x^{2n}\), we need to find the term where \(2k = 2n\), which implies \(k = n\).

### Step 6: Final Calculation
Thus, the coefficient of \(x^{2n}\) in \((1 - x^2)^{2n}\) is:

\[ \binom{2n}{n} (-1)^n \]

### Conclusion
Therefore, the value of the sum is:

\[ S = \binom{2n}{n} (-1)^n \]