Problem Statement
Use the principle of mathematical induction to prove that for all integers $n \ge 1$:
$$\sum_{r=1}^n r(r + 1) = \frac{n(n + 1)(n + 2)}{3}$$
**(a)** [8 marks] Complete the rigorous proof by mathematical induction.
Verified Solution & Marking Scheme
Part (a) [M1]
Base Case $n = 1$: $\text{LHS} = 1(1 + 1) = 2$. $\text{RHS} = \frac{1(2)(3)}{3} = 2$. LHS = RHS, so true for $n = 1$
Part (a) [M1]
Inductive Hypothesis: Assume true for $n = k$, where $k \in \mathbb{Z}^+$: $\sum_{r=1}^k r(r+1) = \frac{k(k+1)(k+2)}{3}$
Part (a) [M1]
Inductive Step: For $n = k + 1$, $\sum_{r=1}^{k+1} r(r+1) = \left[\sum_{r=1}^k r(r+1)\right] + (k+1)(k+2)$
Part (a) [M1]
Substitute hypothesis: $= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$
Part (a) [A1]
Factor out common term $(k+1)(k+2)$: $= (k+1)(k+2)\left[\frac{k}{3} + 1\right] = (k+1)(k+2)\left[\frac{k+3}{3}\right]$
Part (a) [A1]
Rewrite in target form: $= \frac{(k+1)((k+1)+1)((k+1)+2)}{3}$, which is the formula for $n = k + 1$
Part (a) [R1]
Conclusion: Since the statement holds for $n = 1$, and its validity for $n = k$ implies validity for $n = k + 1$, the statement is true for all integers $n \ge 1$ by mathematical induction.