Problem Statement
Prove by mathematical induction that $\sum_{r=1}^{n} r(r!) = (n + 1)! - 1$ for all integers $n \ge 1$.
Verified Solution & Marking Scheme
Base Step: Verify for n = 1
For $n = 1$:
$$\text{LHS} = 1(1!) = 1 \times 1 = 1$$
$$\text{RHS} = (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1$$
Since $\text{LHS} = \text{RHS} = 1$, the proposition is true for $n = 1$.
Inductive Hypothesis: Assume true for n = k
Assume that the statement is true for some positive integer $k \ge 1$:
$$\sum_{r=1}^{k} r(r!) = (k + 1)! - 1$$
Inductive Step: Prove true for n = k + 1
We need to show that $\sum_{r=1}^{k+1} r(r!) = ((k + 1) + 1)! - 1 = (k + 2)! - 1$.
$$\sum_{r=1}^{k+1} r(r!) = \left( \sum_{r=1}^{k} r(r!) \right) + (k + 1)((k + 1)!)$$
Using the inductive hypothesis:
$$= [(k + 1)! - 1] + (k + 1)((k + 1)!)$$
Factor out $(k + 1)!$ from the two terms:
$$= (k + 1)! [1 + (k + 1)] - 1$$
$$= (k + 1)! (k + 2) - 1$$
Since $(k + 1)!(k + 2) = (k + 2)!$:
$$= (k + 2)! - 1 = ((k + 1) + 1)! - 1$$
This matches the required formula for $n = k + 1$.
Formal Conclusion
Since the proposition is true for $n = 1$, and if true for $n = k$, it has been proven true for $n = k + 1$, by the principle of mathematical induction, the statement is true for all integers $n \ge 1$.