The sum `S_(n)` where `T_(n) = (-1)^(n) (n^(2) + n +1)/( n!) ` is

Options

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

Answer

To find the sum \( S_n \) where \( T_n = \frac{(-1)^n (n^2 + n + 1)}{n!} \), we will follow these steps:

### Step 1: Define the Sum
We start with the sum \( S_n \):
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} \frac{(-1)^r (r^2 + r + 1)}{r!} \]

### Step 2: Separate the Terms
We can separate the terms in the summation:
\[ S_n = \sum_{r=1}^{n} \frac{(-1)^r r^2}{r!} + \sum_{r=1}^{n} \frac{(-1)^r r}{r!} + \sum_{r=1}^{n} \frac{(-1)^r}{r!} \]

### Step 3: Simplify Each Summation
We will simplify each of the three summations individually.

1. **For the first term**:
\[ \sum_{r=1}^{n} \frac{(-1)^r r^2}{r!} = \sum_{r=1}^{n} \frac{(-1)^r r}{(r-1)!} = \sum_{r=1}^{n} \frac{(-1)^r r}{r!} + \sum_{r=1}^{n} \frac{(-1)^r}{(r-1)!} \]

2. **For the second term**:
\[ \sum_{r=1}^{n} \frac{(-1)^r r}{r!} = -\sum_{r=0}^{n} \frac{(-1)^r}{(r-1)!} = -\sum_{r=0}^{n} \frac{(-1)^r}{r!} \]

3. **For the third term**:
\[ \sum_{r=1}^{n} \frac{(-1)^r}{r!} \]

### Step 4: Combine the Results
Now we combine the results of the three summations:
\[ S_n = \sum_{r=1}^{n} \frac{(-1)^r r^2}{r!} + \sum_{r=1}^{n} \frac{(-1)^r r}{r!} + \sum_{r=1}^{n} \frac{(-1)^r}{r!} \]

### Step 5: Evaluate the Series
Using the known series expansions:
- The series for \( e^{-x} \) gives us:
\[ \sum_{r=0}^{\infty} \frac{(-1)^r}{r!} = e^{-1} \]
Thus, we can evaluate the sums and find:
\[ S_n = -\frac{(-1)^n}{n!} + \frac{(-1)^{n-1}}{(n-1)!} - \frac{(-1)^n}{n!} \]

### Step 6: Final Result
After simplifying, we find:
\[ S_n = \frac{(-1)^{n-1}}{(n-1)!} - \frac{(-1)^n}{n!} \]

### Conclusion
Thus, the sum \( S_n \) is:
\[ S_n = \frac{(-1)^{n-1}}{(n-1)!} - \frac{(-1)^n}{n!} \]