`sum_(n=0)^(oo) (n^(2) + n + 1)/((n +1)!)` is equal to

Options

A. 2e
B. `2e+1`
C. `2e-1`
D. `1 - 2e`

Answer

To solve the problem, we need to evaluate the infinite series:

\[ S = \sum_{n=0}^{\infty} \frac{n^2 + n + 1}{(n + 1)!} \]

### Step 1: Rewrite the expression
We can rewrite the term \( n^2 + n + 1 \) in a more manageable form:

\[ n^2 + n + 1 = n(n + 1) + 1 \]

Thus, we can express the series as:

\[ S = \sum_{n=0}^{\infty} \frac{n(n + 1)}{(n + 1)!} + \sum_{n=0}^{\infty} \frac{1}{(n + 1)!} \]

### Step 2: Simplify the first term
The first term can be simplified:

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

So we have:

\[ S = \sum_{n=0}^{\infty} \frac{n}{n!} + \sum_{n=0}^{\infty} \frac{1}{(n + 1)!} \]

### Step 3: Evaluate the first summation
The first summation can be evaluated using the property of the exponential function:

\[ \sum_{n=0}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = \sum_{m=0}^{\infty} \frac{1}{m!} = e \]

### Step 4: Evaluate the second summation
The second summation can be rewritten by changing the index:

\[ \sum_{n=0}^{\infty} \frac{1}{(n + 1)!} = \sum_{m=1}^{\infty} \frac{1}{m!} = e - 1 \]

### Step 5: Combine the results
Now we can combine the results from the two summations:

\[ S = e + (e - 1) = 2e - 1 \]

### Final Answer
Thus, the value of the infinite series is:

\[ \boxed{2e - 1} \]

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