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

Options

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

Answer

To solve the summation \( \Sigma_{r=1}^{n} (-1)^{r+1} \frac{nC_r}{r+1} \), we can follow these steps:

### Step 1: Rewrite the summation
We start with the given summation:
\[ S = \sum_{r=1}^{n} (-1)^{r+1} \frac{nC_r}{r+1} \]

### Step 2: Use a combinatorial identity
We can use the identity:
\[ \frac{nC_r}{r+1} = \frac{1}{n+1} \cdot (n+1)C_{r+1} \]
This allows us to rewrite the summation as:
\[ S = \frac{1}{n+1} \sum_{r=1}^{n} (-1)^{r+1} (n+1)C_{r+1} \]

### Step 3: Change the index of summation
To simplify the summation, we change the index of summation by letting \( k = r + 1 \). When \( r = 1 \), \( k = 2 \), and when \( r = n \), \( k = n + 1 \). Thus, we can rewrite the summation:
\[ S = \frac{1}{n+1} \sum_{k=2}^{n+1} (-1)^{k} (n+1)C_k \]

### Step 4: Factor out constants
Now, factor out \( (n+1) \):
\[ S = \frac{1}{n+1} \cdot (n+1) \sum_{k=2}^{n+1} (-1)^{k} C_k \]
This simplifies to:
\[ S = \sum_{k=2}^{n+1} (-1)^{k} C_k \]

### Step 5: Evaluate the summation
The sum \( \sum_{k=0}^{n+1} (-1)^{k} C_k \) is known to be \( 0 \) (due to the binomial theorem). Therefore, we can write:
\[ \sum_{k=0}^{n+1} (-1)^{k} C_k = C_0 + C_1 + \sum_{k=2}^{n+1} (-1)^{k} C_k = 0 \]
This implies:
\[ 1 - 1 + \sum_{k=2}^{n+1} (-1)^{k} C_k = 0 \]
Thus:
\[ \sum_{k=2}^{n+1} (-1)^{k} C_k = 0 \]

### Step 6: Final result
Putting it all together, we find:
\[ S = 0 \]

### Conclusion
The value of the summation \( \Sigma_{r=1}^{n} (-1)^{r+1} \frac{nC_r}{r+1} \) is:
\[ \boxed{0} \]