The value of `sum_(r=1)^(n)(-1)^(r-1)((r )/(r+1))*^(n)C_(r )` is (a) `1/(n+1)` (b) `1/n` (c) `1/(n-1)` (d) 0

Options

A. `(1)/(n+1)`
B. `(1)/(n)`
C. `(1)/(n-1)`
D. `0`

Explanation

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

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

### Step 1: Rewrite the term \(\frac{r}{r+1}\)

We can rewrite \(\frac{r}{r+1}\) as follows:

\[ \frac{r}{r+1} = 1 - \frac{1}{r+1} \]

### Step 2: Substitute into the summation

Substituting this back into the summation gives us:

\[ \sum_{r=1}^{n} (-1)^{r-1} \left(1 - \frac{1}{r+1}\right) \binom{n}{r} \]

This can be split into two separate summations:

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

### Step 3: Evaluate the first summation

The first summation is:

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

This is equal to:

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

### Step 4: Evaluate the second summation

Now we need to evaluate:

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

Using the identity \(\frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} \binom{n+1}{r+1}\), we can rewrite the second summation as:

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

By changing the index of summation, we can let \(k = r + 1\), which gives us:

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

This can be simplified to:

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

### Step 5: Combine the results

Combining both parts, we have:

\[ 1 - \frac{1}{n+1} \cdot 1 = 1 - \frac{1}{n+1} = \frac{n}{n+1} \]

### Final Result

Thus, the value of the original summation is:

\[ \frac{1}{n+1} \]

### Conclusion

The final answer is:

\[ \boxed{\frac{1}{n+1}} \]