Problem Statement
Mathematical induction is a cornerstone of formal analysis in Higher Level Mathematics.
**(a)** [7 marks] Use mathematical induction to prove that for all positive integers $n \in \mathbb{Z}^+$:
$$\sum_{r=1}^{n} r \cdot r! = (n + 1)! - 1$$
Verified Solution & Marking Scheme
Part (a) [M1]
Base Case ($n = 1$): LHS $= 1 \cdot 1! = 1$. RHS $= (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1$. Since LHS $=$ RHS, the statement is true for $n = 1$.
Part (a) [M1]
Inductive Hypothesis: Assume the statement is true for some positive integer $n = k$, so $\sum_{r=1}^{k} r \cdot r! = (k + 1)! - 1$.
Part (a) [M1]
Inductive Step ($n = k + 1$): Consider $\sum_{r=1}^{k+1} r \cdot r! = \sum_{r=1}^{k} r \cdot r! + (k + 1)(k + 1)!$
Part (a) [A1]
Substitute the induction hypothesis: $= (k + 1)! - 1 + (k + 1)(k + 1)!$
Part (a) [A1]
Factor out $(k + 1)!$: $= (k + 1)![1 + (k + 1)] - 1 = (k + 1)!(k + 2) - 1$
Part (a) [A1]
Recognize that $(k + 2)(k + 1)! = (k + 2)! = ((k + 1) + 1)! - 1$, which is the statement for $n = k + 1$.
Part (a) [R1]
Formal conclusion: Since the statement is true for $n = 1$, and if true for $n = k$ then it is proven true for $n = k + 1$, by the principle of mathematical induction the statement holds true for all $n \in \mathbb{Z}^+$.