Problem Statement
Use the principle of mathematical induction to prove that $\sum_{r=1}^n r \cdot r! = (n+1)! - 1$ for all integers $n \ge 1$.
**(a)** [7 marks] Prove by induction that the formula holds for all positive integers $n$.
Verified Solution & Marking Scheme
Part (a) [M1]
Base Case ($n = 1$): LHS $= 1 \cdot 1! = 1$. RHS $= (1+1)! - 1 = 2! - 1 = 1$. LHS = RHS, true for $n = 1$.
Part (a) [M1]
Inductive Hypothesis: Assume true for $n = k$, i.e., $\sum_{r=1}^k r \cdot r! = (k+1)! - 1$.
Part (a) [M1]
Inductive Step: For $n = k+1$, $\sum_{r=1}^{k+1} r \cdot r! = \left(\sum_{r=1}^k r \cdot r!\right) + (k+1)(k+1)!$
Part (a) [A1]
$= (k+1)! - 1 + (k+1)(k+1)! = (k+1)![1 + (k+1)] - 1$
Part (a) [A1]
$= (k+1)!(k+2) - 1 = (k+2)! - 1 = ((k+1)+1)! - 1$
Part (a) [R1]
Conclusion: Since true for $n = 1$ and true for $n = k+1$ whenever true for $n = k$, the statement is true for all $n \in \mathbb{Z}^+$ by mathematical induction.
Part (a) [Note]
Award R1 for explicit statement: "Assume true for $n = k$, i.e., $3^{2k} - 1 = 8M$ for $M \in \mathbb{Z}$".
Part (a) [Note]
Award R1 for full concluding statement linking base case $n=1$ and induction step $n=k \implies n=k+1$.