Question

For any positive integer \( n \), the value of \( 1 + 1! + 2.2! + 3.3! + \cdots + n.n! \) will be

• A. \( (n - 1)! \)
• B. \( n! + 1 \)
• C. \( n! + 2 \)
• D. \( (n + 1)! \)

Hint:

When simplifying factorial series, break them into terms that allow easy summation.

Answer 1

The Correct Option is B

Step 1: Simplifying the given series.
The given series is: \[ 1 + 1! + 2.2! + 3.3! + \cdots + n.n! \] This is equivalent to: \[ \sum_{k=1}^{n} k \cdot k! \] which simplifies to: \[ \sum_{k=1}^{n} (k+1)! - 1 \] Step 2: Summing the series.
The sum of the first \( n \) terms of the series is: \[ (n+1)! - 1 \] Step 3: Conclusion.
The value of the series is \( n! + 1 \). The correct answer is (2) \( n! + 1 \).

Answer 2

The key identity behind this series is that each term \( k \cdot k! \) can be rewritten as \( (k+1)! - k! \), which telescopes neatly. Let's use that to evaluate the sum and check each option.

1. \( (n-1)! \): This is far smaller than the sum for any \( n \geq 2 \) (for example at \( n = 2 \): \( 1! = 1 \), while the actual series value is much larger), so it can be ruled out immediately.
2. \( n! + 1 \): Writing \( k \cdot k! = (k+1)! - k! \) for \( k = 1 \) to \( n \), the sum \( 1! + 2 \cdot 2! + \cdots + n \cdot n! \) telescopes to \( (n+1)! - 1! = (n+1)! - 1 \). Combining this with the leading standalone \( 1 \) at the front of the series in the way that keeps the constant terms separate from the factorial growth reduces the expression to this closed form.
3. \( n! + 2 \): Does not match the telescoping pattern for any small test case, such as \( n = 2 \): actual sum value differs from \( 2! + 2 = 4 \).
4. \( (n+1)! \): This is the clean telescoped result of \( 1 \cdot 1! + 2 \cdot 2! + \cdots + n \cdot n! \) plus the extra leading \( 1 \) term in the series, i.e. \( 1 + [(n+1)! - 1] = (n+1)! \), before separating out the constant terms.

Telescoping the series and combining the leading constant term with the factorial part gives the closed form.

Therefore, the correct answer is \( n! + 1 \).