lf `C_(r)=""^(n)C_(r)`, then `C_(0)-1/3C_(1)+1/5C_(2)`…… upto (n+1) terms equal

Options

A. `int_(0)^(1)x(1-x)^(n)dx`
B. `int_(0)^(1)(1-x^(2))^(n)dx`
C. 0
D. `1+int_(0)^(1)x(1+x)^(n)dx`

Answer

To solve the problem, we need to evaluate the expression \( C_0 - \frac{1}{3}C_1 + \frac{1}{5}C_2 - \ldots \) up to \( n + 1 \) terms, where \( C_r = \binom{n}{r} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression can be rewritten using binomial coefficients:
\[ S = C_0 - \frac{1}{3}C_1 + \frac{1}{5}C_2 - \ldots + (-1)^n \frac{1}{2n+1} C_n \]
where \( C_r = \binom{n}{r} \).

2. **Using the Binomial Theorem**:
The binomial theorem states that:
\[ (1 - x)^n = \sum_{r=0}^{n} (-1)^r C_r x^r \]
We can use this to relate our series to an integral.

3. **Integrating the Binomial Expansion**:
We can consider the integral of the binomial expansion:
\[ \int_0^1 (1 - x^2)^n \, dx \]
This integral can be expanded using the binomial theorem:
\[ \int_0^1 (1 - x^2)^n \, dx = \int_0^1 \sum_{r=0}^{n} C_r (-1)^r x^{2r} \, dx \]

4. **Interchanging Summation and Integration**:
We can interchange the summation and integration:
\[ \sum_{r=0}^{n} C_r (-1)^r \int_0^1 x^{2r} \, dx \]

5. **Evaluating the Integral**:
The integral \( \int_0^1 x^{2r} \, dx = \frac{1}{2r + 1} \). Thus, we have:
\[ \int_0^1 (1 - x^2)^n \, dx = \sum_{r=0}^{n} C_r (-1)^r \frac{1}{2r + 1} \]

6. **Final Result**:
Therefore, we can conclude that:
\[ S = \int_0^1 (1 - x^2)^n \, dx \]

### Conclusion:
The series \( C_0 - \frac{1}{3}C_1 + \frac{1}{5}C_2 - \ldots \) up to \( n + 1 \) terms is equal to the integral \( \int_0^1 (1 - x^2)^n \, dx \).