Step-by-Step Solution
Key Concept: Recognizing that the expression telescopes when rewritten as a difference of consecutive factorial-like terms.
The sum $\sum_{r=1}^{99} r[(r+1)^2 - r] = \sum_{r=1}^{99} r[r^2 + 2r + 1 - r] = \sum_{r=1}^{99} r(r^2 + r + 1) = \sum_{r=1}^{99} [(r+1)(r+1)! - r!] $ telescopes. This evaluates to $100! - 1 = 100(100!) - 1$.
Correct Answer: 2