`sum_(r=1)^n (r^2 - r-1)/((r+1)!)` is equal to :

Options

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

Answer

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

\[ S = \sum_{r=1}^{n} \frac{r^2 - r - 1}{(r+1)!} \]

### Step 1: Simplifying the Numerator
We start with the expression in the numerator, \( r^2 - r - 1 \). We can manipulate this expression by adding and subtracting \( r \):

\[ r^2 - r - 1 = r^2 + r - 2r - 1 = (r^2 + r) - (2r + 1) \]

### Step 2: Rewriting the Sum
Now, we can rewrite the sum \( S \):

\[ S = \sum_{r=1}^{n} \frac{(r^2 + r) - (2r + 1)}{(r+1)!} \]

This can be separated into two sums:

\[ S = \sum_{r=1}^{n} \frac{r^2 + r}{(r+1)!} - \sum_{r=1}^{n} \frac{2r + 1}{(r+1)!} \]

### Step 3: Simplifying Each Term
Now we will simplify each term separately.

1. **First Term:**
\[ \sum_{r=1}^{n} \frac{r^2 + r}{(r+1)!} = \sum_{r=1}^{n} \frac{r(r+1)}{(r+1)!} = \sum_{r=1}^{n} \frac{r}{r!} = \sum_{r=1}^{n} \frac{1}{(r-1)!} \]
This sum can be rewritten as:
\[ \sum_{r=0}^{n-1} \frac{1}{r!} = e - \frac{1}{n!} \]

2. **Second Term:**
\[ \sum_{r=1}^{n} \frac{2r + 1}{(r+1)!} = 2\sum_{r=1}^{n} \frac{r}{(r+1)!} + \sum_{r=1}^{n} \frac{1}{(r+1)!} \]
The first part can be simplified:
\[ 2\sum_{r=1}^{n} \frac{r}{(r+1)!} = 2\sum_{r=1}^{n} \frac{1}{r!} - 2\sum_{r=1}^{n} \frac{1}{(r+1)!} = 2\left(e - \frac{1}{n!}\right) - 2\left(e - \frac{1}{(n+1)!}\right) \]

The second part simplifies to:
\[ \sum_{r=1}^{n} \frac{1}{(r+1)!} = \sum_{r=2}^{n+1} \frac{1}{r!} = e - 1 - \frac{1}{(n+1)!} \]

### Step 4: Combining the Results
Now we combine the results from both sums:

\[ S = \left(e - \frac{1}{n!}\right) - \left(2\left(e - \frac{1}{n!}\right) - 2\left(e - \frac{1}{(n+1)!}\right) + e - 1 - \frac{1}{(n+1)!}\right) \]

### Step 5: Final Simplification
After simplification, we can find the final expression for \( S \):

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

### Conclusion
Thus, the sum \( S \) is equal to:

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

### Final Answer
The final answer is:

\[ -\frac{1}{(n+1) \cdot n!} \]