Standard +0.8 This is a standard mathematical induction proof with a summation involving factorials. While the algebraic manipulation requires recognizing that r × r! = (r+1)! - r!, the proof structure is routine for Further Maths students. The question is moderately challenging due to the factorial manipulation needed, but follows a well-practiced template, placing it somewhat above average difficulty.
Prove by mathematical induction that $\sum_{r=1}^{n}(r \times r!) = (n + 1)! - 1$ for all positive integers $n$. [6]
\hfill \mbox{\textit{SPS SPS FM Pure 2025 Q2 [6]}}