Sum of the series `sum _(r=1) ^n (r^2+1) (r!)` is

Options

A. `(n+1)!`
B. `(n+2)!-1`
C. `n(n+1)!`
D. none of these

Answer

To solve the series \( \sum_{r=1}^{n} (r^2 + 1) (r!) \), we will evaluate it step by step.

### Step 1: Write down the series
We are given the series:
\[ S_n = \sum_{r=1}^{n} (r^2 + 1) (r!) \]

### Step 2: Break down the expression
The expression \( (r^2 + 1) (r!) \) can be split into two parts:
\[ S_n = \sum_{r=1}^{n} (r^2 \cdot r!) + \sum_{r=1}^{n} (1 \cdot r!) \]
This simplifies to:
\[ S_n = \sum_{r=1}^{n} (r^2 \cdot r!) + \sum_{r=1}^{n} (r!) \]

### Step 3: Calculate \( \sum_{r=1}^{n} (r!) \)
The sum of factorials can be calculated directly:
\[ \sum_{r=1}^{n} (r!) = 1! + 2! + 3! + \ldots + n! \]

### Step 4: Calculate \( \sum_{r=1}^{n} (r^2 \cdot r!) \)
To calculate \( \sum_{r=1}^{n} (r^2 \cdot r!) \), we can use the identity \( r^2 \cdot r! = r \cdot (r \cdot r!) = r \cdot (r+1)! - r \cdot r! \):
\[ \sum_{r=1}^{n} (r^2 \cdot r!) = \sum_{r=1}^{n} (r \cdot (r+1)!) - \sum_{r=1}^{n} (r \cdot r!) \]

### Step 5: Simplify the sums
Using the above identity, we can express the sums:
1. The first part becomes:
\[ \sum_{r=1}^{n} (r \cdot (r+1)!) = n \cdot (n+1)! - 1! \]
2. The second part is:
\[ \sum_{r=1}^{n} (r \cdot r!) = n! + (n-1)! + \ldots + 1! \]

### Step 6: Combine the results
Now we can combine the results:
\[ S_n = \sum_{r=1}^{n} (r^2 \cdot r!) + \sum_{r=1}^{n} (r!) = (n \cdot (n+1)! - 1!) + (1! + 2! + \ldots + n!) \]

### Step 7: Final expression
After simplification, we find:
\[ S_n = n \cdot (n+1)! + \sum_{r=1}^{n} (r!) \]

### Conclusion
The final result for the series \( S_n \) can be expressed as:
\[ S_n = n \cdot (n+1)! + \sum_{r=1}^{n} (r!) \]