If n is odd, then sum of the series `C_(0)^(2)-C_(1)^(2)+C_(2)^(2)-C_(2)^(3)+…..+(-1)^(n)C_(n)^(2)` is

Options

A. 0
B. 1
C. `""^(2n)C_(n)`
D. `-(""^(2n)C_(n))`

Answer

To solve the problem, we need to evaluate the sum of the series given by:

\[ S = C_0^2 - C_1^2 + C_2^2 - C_3^2 + \ldots + (-1)^n C_n^2 \]

where \( n \) is an odd integer.

### Step 1: Understand the Binomial Coefficient
The binomial coefficient \( C_r^n \) (or \( \binom{n}{r} \)) represents the number of ways to choose \( r \) elements from a set of \( n \) elements. The important property we will use is:

\[ C_r^n = C_{n-r}^n \]

### Step 2: Write the Series
The series can be rewritten as:

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

### Step 3: Use the Binomial Theorem
According to the Binomial Theorem, we have:

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

If we substitute \( x = -1 \), we get:

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

### Step 4: Evaluate the Left Side
Since \( n \) is odd, we have:

\[ 0^n = 0 \]

Thus, we find:

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

### Step 5: Conclusion
This implies that:

\[ S = 0 \]

Therefore, the sum of the series is:

\[ \boxed{0} \]