If `a_n = n (n!)`, then `sum_(r=1)^100 a_r` is equal to

Options

A. 101!
B. 100!-1
C. 101!-1
D. 101!+1

Explanation

To solve the problem, we need to find the sum \( \sum_{r=1}^{100} a_r \) where \( a_n = n \cdot n! \).

### Step-by-step Solution:

1. **Define \( a_r \)**:
Given \( a_n = n \cdot n! \), we can express \( a_r \) as:
\[ a_r = r \cdot r! \]

2. **Rewrite \( a_r \)**:
We can manipulate \( a_r \) as follows:
\[ a_r = r \cdot r! = (r + 1 - 1) \cdot r! = (r + 1) \cdot r! - r! \]
This simplifies to:
\[ a_r = (r + 1)! - r! \]

3. **Set up the summation**:
Now we can set up the summation:
\[ \sum_{r=1}^{100} a_r = \sum_{r=1}^{100} \left( (r + 1)! - r! \right) \]

4. **Expand the summation**:
Expanding the summation gives:
\[ \sum_{r=1}^{100} a_r = \left( 2! - 1! \right) + \left( 3! - 2! \right) + \left( 4! - 3! \right) + \ldots + \left( 101! - 100! \right) \]

5. **Observe the telescoping nature**:
Notice that this is a telescoping series. Most terms will cancel out:
\[ = 101! - 1! \]

6. **Calculate the final result**:
Since \( 1! = 1 \), we have:
\[ \sum_{r=1}^{100} a_r = 101! - 1 \]

### Final Answer:
Thus, the sum \( \sum_{r=1}^{100} a_r \) is:
\[ \boxed{101! - 1} \]