Question

$\sum_{r=1}^{n} (r \cdot r!) = $ ________

• A. $1$
• B. $n$
• C. $(n+1)! - 1$
• D. $0$

Hint:

For series summation problems, especially those involving factorials or fractions, always look for a way to decompose the general term into a difference $f(r+1) - f(r)$. This leads to a telescoping series, which is easily evaluated. The identity $r \cdot r! = (r+1)! - r!$ is a classic trick.

Answer 1

The Correct Option is C

Step 1: Write the general term
Let \[ t_r = r \cdot r! \]
Step 2: Transform the term
Rewrite $r$ as $(r+1 - 1)$: \[ t_r = (r+1 - 1)\cdot r! \] \[ t_r = (r+1)r! - r! \]
Step 3: Use factorial property
\[ (r+1)r! = (r+1)! \] So, \[ t_r = (r+1)! - r! \]
Step 4: Form the series
\[ \sum_{r=1}^{n} r \cdot r! = \sum_{r=1}^{n} \big[(r+1)! - r!\big] \]
Step 5: Expand the series
\[ = (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + ((n+1)! - n!) \]
Step 6: Apply telescoping
All intermediate terms cancel: \[ = -1! + (n+1)! \]
Step 7: Simplify
\[ = (n+1)! - 1 \] Final Answer:
\[ \boxed{(n+1)! - 1} \]