The value of `sum_(r=1)^(n) (-1)^(r+1)(""^(n)C_(r))/(r+1)` is equal to

Options

A. (a) `-1/(n+1)`
B. (b) `-(1)/(n)`
C. (c) `(1)/(n+1)`
D. (d) `(n)/(n+1)`

Explanation

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

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

### Step 1: Rewrite the Summation
We can rewrite the summation \( S \) as:

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

### Step 2: Use the Binomial Theorem
Recall that the Binomial Theorem states:

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

### Step 3: Integrate Both Sides
Integrate both sides with respect to \( x \):

\[ \int (1 - x)^n \, dx = \int \sum_{r=0}^{n} \binom{n}{r} (-x)^r \, dx \]

The left-hand side becomes:

\[ -\frac{(1-x)^{n+1}}{n+1} + C \]

The right-hand side becomes:

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

### Step 4: Evaluate the Integral from 0 to 1
Evaluate the integral from \( 0 \) to \( 1 \):

\[ \int_0^1 (1 - x)^n \, dx = \left[-\frac{(1-x)^{n+1}}{n+1}\right]_0^1 = \frac{1}{n+1} \]

### Step 5: Interchange Summation and Integration
Interchange the summation and integration:

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

### Step 6: Separate the r=0 Term
Separate the term for \( r=0 \):

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

### Step 7: Solve for S
Now we have:

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

Thus,

\[ S = \frac{1}{n+1} - 1 = \frac{1 - (n+1)}{n+1} = \frac{-n}{n+1} \]

### Step 8: Final Result
Therefore, the value of the summation is:

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