If `C_(r)` stands for `""^(n)C_(r)`, then the sum of the series `(2((n)/(2))!((n)/(2))!)/(n!)[C_(0)^(2)-2C_(1)^(2)+3C_(2)^(2)-...+(-1)^(n)(n+1)C_(n)^(2)]`, where n is an even positive integers, is:

Options

A. `(-1)^(n//2)(n+2)`
B. `(-1)^(n)(n+1)`
C. `(-1)^(n//2)(n+1)`
D. None of these

Answer

To solve the given problem, we will follow the steps outlined in the video transcript and derive the solution systematically.

### Step-by-Step Solution:

1. **Understanding the Series**:
We need to evaluate the series:
\[ C(0)^2 - 2C(1)^2 + 3C(2)^2 - \ldots + (-1)^n(n+1)C(n)^2 \]
where \( C(r) = \binom{n}{r} \).

2. **Using Binomial Expansion**:
We start with the binomial expansion of \( (1+x)^n \):
\[ (1+x)^n = C(0) + C(1)x + C(2)x^2 + \ldots + C(n)x^n \]

3. **Multiplying by \( x \)**:
Multiply both sides by \( x \):
\[ x(1+x)^n = C(0)x + C(1)x^2 + C(2)x^3 + \ldots + C(n)x^{n+1} \]

4. **Differentiating**:
Differentiate both sides with respect to \( x \):
\[ n(1+x)^{n-1} + (1+x)^n = C(0) + 2C(1)x + 3C(2)x^2 + \ldots + (n+1)C(n)x^n \]

5. **Rearranging**:
Rearranging gives us:
\[ n(1+x)^{n-1} + (1+x)^n = C(0) + 2C(1)x + 3C(2)x^2 + \ldots + (n+1)C(n)x^n \]

6. **Introducing a New Term**:
Introduce \( 1 - \frac{1}{x} \):
\[ C(0) - \frac{C(1)}{x} + \frac{C(2)}{x^2} - \ldots + \frac{C(n)}{x^n} \]

7. **Combining Terms**:
Multiply the two expressions obtained from differentiation and rearranging:
\[ (1+x)^{n-1}(x - 1) = x \cdot (n+1)C(n)x^n \]

8. **Finding the Independent Term**:
We need to find the term independent of \( x \) in the resulting expression.

9. **Evaluating the Coefficients**:
The coefficients will give us the required sums:
\[ \sum_{r=0}^{n} (-1)^r (r+1)C(r)^2 \]

10. **Final Expression**:
After simplifying, we find:
\[ \frac{2}{nC(n/2)} \left( (-1)^{n/2}(n+2) \right) \]

11. **Final Answer**:
Thus, the sum of the series is:
\[ \boxed{(-1)^{n/2}(n+2)} \]