Problem Statement
Use the principle of mathematical induction to prove that:
$$\sum_{r=1}^{n} r(r!) = (n+1)! - 1$$
for all $n \in \mathbb{Z}^+$.
Verified Solution & Marking Scheme
Base Case: n = 1
LHS $= \sum_{r=1}^{1} r(r!) = 1(1!) = 1$.
RHS $= (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1$.
Since LHS $=$ RHS, the statement is true for $n = 1$.
Inductive Hypothesis: Assume true for n = k
Assume the statement holds for $n = k$, where $k \in \mathbb{Z}^+$:
$$\sum_{r=1}^{k} r(r!) = (k+1)! - 1$$
Inductive Step: Prove for n = k + 1
For $n = k + 1$:
$$\sum_{r=1}^{k+1} r(r!) = \left[ \sum_{r=1}^{k} r(r!) \right] + (k+1)(k+1)!$$
Using the induction hypothesis:
$$= [(k+1)! - 1] + (k+1)(k+1)!$$
Group terms with $(k+1)!$:
$$= (k+1)!(1 + k + 1) - 1 = (k+1)!(k + 2) - 1$$
Since $(k+1)!(k+2) = (k+2)! = ((k+1) + 1)!$:
$$= ((k+1) + 1)! - 1$$
Formal Inductive Conclusion
Since the proposition is true for $n = 1$, and if true for $n = k$ it is also true for $n = k + 1$, then by the principle of mathematical induction, the proposition is true for all $n \in \mathbb{Z}^+$.