Problem Statement
Prove by mathematical induction that for all positive integers $n \in \mathbb{Z}^+$:
$\sum_{r=1}^{n} r(r + 1) = \frac{n(n + 1)(n + 2)}{3}$
Verified Solution & Marking Scheme
Base Step (n = 1)
For $n = 1$:
$\text{LHS} = 1(1 + 1) = 2$
$\text{RHS} = \frac{1(1 + 1)(1 + 2)}{3} = \frac{1(2)(3)}{3} = 2$
Since $\text{LHS} = \text{RHS}$, the statement is true for $n = 1$.
Induction Hypothesis
Assume the statement is true for $n = k$, where $k \in \mathbb{Z}^+$:
$\sum_{r=1}^{k} r(r + 1) = \frac{k(k + 1)(k + 2)}{3}$
Inductive Step (n = k + 1)
We must show that:
$\sum_{r=1}^{k+1} r(r + 1) = \frac{(k + 1)(k + 2)(k + 3)}{3}$
Starting from LHS:
$\sum_{r=1}^{k+1} r(r + 1) = \sum_{r=1}^{k} r(r + 1) + (k + 1)(k + 2)$
$= \frac{k(k + 1)(k + 2)}{3} + (k + 1)(k + 2) = (k + 1)(k + 2) \left[ \frac{k}{3} + 1 \right] = (k + 1)(k + 2) \left[ \frac{k + 3}{3} \right] = \frac{(k + 1)(k + 2)(k + 3)}{3}$
Conclusion
Since the statement is true for $n = 1$, and its truth for $n = k$ implies its truth for $n = k + 1$, the statement is true for all $n \in \mathbb{Z}^+$ by the Principle of Mathematical Induction.