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

To solve the given problem, we need to evaluate the sum $\sum_{r=1}^{n} (r \cdot r!)$. Let's explore the steps to arrive at the solution.

The given sum can be expanded as:

• $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + n \cdot n!$

We observe that each term in the form $r \cdot r!$ can be rewritten using a simple algebraic manipulation:

• $r \cdot r! = r \cdot r \cdot (r-1)! = (r+1-1) \cdot r! = (r+1)! - r!$

This indicates that each term $r \cdot r!$ can be expressed as the difference of two factorials. Thus, the sum becomes a telescopic series:

• $\sum_{r=1}^{n} (r \cdot r!) = [(2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + ((n+1)! - n!)]$

Observe the cancellation pattern in this telescopic series:

• The terms except for $1!$ and $(n+1)!$ cancel each other out.
• This leaves us with: $(n+1)! - 1!$

Since $1! = 1$, the final expression simplifies to:

• $(n+1)! - 1$

Therefore, the evaluated sum is $(n+1)! - 1$, which matches the provided correct answer.

Conclusion: The correct answer is $(n+1)! - 1$.