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. `n*(n+2)!`

Answer

To find the sum of the series \( \sum_{r=1}^{n} (r^2 + 1) r! \), we can break it down step by step.

### Step 1: Split the Series
We can rewrite the series as:
\[ \sum_{r=1}^{n} (r^2 + 1) r! = \sum_{r=1}^{n} r^2 r! + \sum_{r=1}^{n} r! \]

### Step 2: Simplify the First Part
The first part, \( \sum_{r=1}^{n} r^2 r! \), can be rewritten using the identity \( r^2 = (r-1)(r+1) + 1 \):
\[ \sum_{r=1}^{n} r^2 r! = \sum_{r=1}^{n} ((r-1)(r+1) + 1) r! = \sum_{r=1}^{n} (r-1)(r+1) r! + \sum_{r=1}^{n} r! \]

### Step 3: Rewrite the First Part
Now, we can express \( (r-1)(r+1) r! \) as:
\[ (r-1)(r+1) r! = (r-1) (r+1)! = (r+1)! - (r-1)! \]
Thus, we have:
\[ \sum_{r=1}^{n} r^2 r! = \sum_{r=1}^{n} ((r+1)! - (r-1)!) + \sum_{r=1}^{n} r! \]

### Step 4: Evaluate the Series
Now we can evaluate the series:
\[ \sum_{r=1}^{n} ((r+1)! - (r-1)!) = (2! - 0!) + (3! - 1!) + (4! - 2!) + \ldots + ((n+1)! - (n-1)!) \]
This telescopes to:
\[ (n+1)! - 0! + 1! + 2! + \ldots + n! \]

### Step 5: Combine the Results
Now, we combine everything:
\[ \sum_{r=1}^{n} (r^2 + 1) r! = (n+1)! + \sum_{r=1}^{n} r! \]
The sum of factorials \( \sum_{r=1}^{n} r! \) is a known series.

### Step 6: Final Expression
Thus, the final expression for the sum of the series is:
\[ \sum_{r=1}^{n} (r^2 + 1) r! = (n+1)! + \sum_{r=1}^{n} r! \]