If n is an odd natural number, then `sum_(r=0)^(n) ((-1)^(r ))/(""^(n)C_(r ))` equals

Options

A. 0
B. `(1)/(n)`
C. `(n)/(2^(n))`
D. none of these

Answer

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

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

where \( n \) is an odd natural number.

### Step 1: Write down the sum explicitly
First, we can write the sum explicitly for \( n \) being an odd natural number:

\[ S = \frac{(-1)^0}{\binom{n}{0}} + \frac{(-1)^1}{\binom{n}{1}} + \frac{(-1)^2}{\binom{n}{2}} + \cdots + \frac{(-1)^n}{\binom{n}{n}} \]

### Step 2: Identify the terms
Since \( n \) is odd, we can pair the terms in the sum. The terms can be grouped as follows:

- The first term is \( \frac{1}{\binom{n}{0}} \)
- The second term is \( -\frac{1}{\binom{n}{1}} \)
- The third term is \( \frac{1}{\binom{n}{2}} \)
- The fourth term is \( -\frac{1}{\binom{n}{3}} \)
- ...
- The last term is \( -\frac{1}{\binom{n}{n}} \)

### Step 3: Pair the terms
We can pair the terms as follows:

\[ S = \left( \frac{1}{\binom{n}{0}} - \frac{1}{\binom{n}{1}} \right) + \left( \frac{1}{\binom{n}{2}} - \frac{1}{\binom{n}{3}} \right) + \cdots + \left( \frac{1}{\binom{n}{n-1}} - \frac{1}{\binom{n}{n}} \right) \]

### Step 4: Use the property of binomial coefficients
We know that:

\[ \binom{n}{r} = \binom{n}{n-r} \]

Using this property, we can rewrite the terms in pairs:

\[ S = \left( \frac{1}{\binom{n}{0}} - \frac{1}{\binom{n}{1}} \right) + \left( \frac{1}{\binom{n}{2}} - \frac{1}{\binom{n}{3}} \right) + \cdots + \left( \frac{1}{\binom{n}{(n-1)/2}} - \frac{1}{\binom{n}{(n+1)/2}} \right) \]

### Step 5: Evaluate the sum
Notice that for each pair, the terms will cancel out because:

\[ \frac{1}{\binom{n}{r}} + \frac{1}{\binom{n}{n-r}} = 0 \]

Thus, all pairs will cancel out, leading to:

\[ S = 0 \]

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

\[ \boxed{0} \]