`sum_(n=1)^(oo)(2 n^2+n+1)/((n!))` =

Options

A. 2e - 1
B. 2e + 1
C. 6e - 1
D. 6e + 1

Answer

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

\[ \sum_{n=1}^{\infty} \frac{2n^2 + n + 1}{n!} \]

### Step-by-Step Solution:

1. **Rewrite the series:**
We can express the series as:
\[ \sum_{n=1}^{\infty} \frac{2n^2 + n + 1}{n!} = \sum_{n=1}^{\infty} \frac{2n^2}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} + \sum_{n=1}^{\infty} \frac{1}{n!} \]

2. **Evaluate each component:**
- **First term:**
\[ \sum_{n=1}^{\infty} \frac{2n^2}{n!} = 2 \sum_{n=1}^{\infty} \frac{n^2}{n!} \]
We know that:
\[ \sum_{n=0}^{\infty} \frac{n^2}{n!} = e \]
Therefore:
\[ \sum_{n=1}^{\infty} \frac{n^2}{n!} = e \]
Thus:
\[ 2 \sum_{n=1}^{\infty} \frac{n^2}{n!} = 2e \]

- **Second term:**
\[ \sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=0}^{\infty} \frac{n}{n!} = e \]

- **Third term:**
\[ \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \]

3. **Combine the results:**
Now we combine all the evaluated components:
\[ \sum_{n=1}^{\infty} \frac{2n^2 + n + 1}{n!} = 2e + e + (e - 1) \]
Simplifying this gives:
\[ = 2e + e + e - 1 = 5e - 1 \]

4. **Final Result:**
Therefore, the sum of the series is:
\[ \sum_{n=1}^{\infty} \frac{2n^2 + n + 1}{n!} = 5e - 1 \]