`sum_(r = 1)^(n) r. r!` is equal to

Options

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

Answer

To find the value of the summation \( \sum_{r=1}^{n} r \cdot r! \), we can follow these steps:

### Step 1: Rewrite the term \( r \cdot r! \)
We start by rewriting the term \( r \cdot r! \) in a more manageable form. Notice that:
\[ r \cdot r! = r! \cdot r = (r + 1 - 1) \cdot r! = (r + 1)! - r! \]
This means we can express \( r \cdot r! \) as the difference of two factorials.

### Step 2: Substitute into the summation
Now, we can substitute this expression back into our summation:
\[ \sum_{r=1}^{n} r \cdot r! = \sum_{r=1}^{n} ((r + 1)! - r!) \]

### Step 3: Expand the summation
Next, we can expand the summation:
\[ \sum_{r=1}^{n} ((r + 1)! - r!) = \sum_{r=1}^{n} (r + 1)! - \sum_{r=1}^{n} r! \]

### Step 4: Evaluate the sums
Now, let's evaluate each sum separately:
- The first sum \( \sum_{r=1}^{n} (r + 1)! \) can be rewritten by changing the index of summation:
\[ \sum_{r=1}^{n} (r + 1)! = \sum_{s=2}^{n+1} s! \quad \text{(where \( s = r + 1 \))} \]
This sum is \( 2! + 3! + 4! + \ldots + (n + 1)! \).

- The second sum \( \sum_{r=1}^{n} r! \) is simply \( 1! + 2! + 3! + \ldots + n! \).

### Step 5: Combine the results
Now, we can combine the results:
\[ \sum_{s=2}^{n+1} s! - \sum_{r=1}^{n} r! = (2! + 3! + \ldots + (n + 1)!) - (1! + 2! + 3! + \ldots + n!) \]
Notice that all terms from \( 2! \) to \( n! \) will cancel out, leaving us with:
\[ (n + 1)! - 1! \]
Thus, we have:
\[ \sum_{r=1}^{n} r \cdot r! = (n + 1)! - 1 \]

### Final Answer
Therefore, the final result is:
\[ \sum_{r=1}^{n} r \cdot r! = (n + 1)! - 1 \]

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