AHL 1.15—Proof by induction, contradiction, counterexamples

The base case requires verifying the formula at $n=1$. We check whether $1^3=\frac{1^2\cdot 2^2}{4}$.

LHS $=1^3=1$.

RHS $=\frac{1^2\cdot (1+1{)}^2}{4}=\frac{1\cdot 4}{4}=1$.

Since LHS $=1=$ RHS, $P(1)$ holds. The correct option states exactly this: LHS $=1$, RHS $=\frac{1\cdot 4}{4}=1$.

We need the correct base case verification — the right formula at $n=1$ with both sides computed accurately.

Option B gives RHS $=\frac{1\cdot 2}{2}=1$, but this corresponds to the formula $\frac{n(n+1)}{2}$, which is the sum of integers, not cubes. Wrong formula used.

Option C claims RHS $=2$ at $n=1$, but $\frac{1^2\cdot 2^2}{4}=\frac{4}{4}=1\ne 2$. This contains an arithmetic error.

Option D uses $n=2$ as the base case, computing $1^3+2^3=9$. But the domain begins at $n=1$, so $n=2$ is not the correct base case.

Option A correctly computes both sides at $n=1$: LHS $=1$ and RHS $=\frac{1\cdot 4}{4}=1$, confirming $P(1)$ holds.

The inductive hypothesis is the assumption that $P(k)$ is true — that is, the formula holds when $n=k$.

Replace $n$ with $k$ in both the sum and the closed form: $\sum _{r=1}^{k}r(r+1)=\frac{k(k+1)(k+2)}{3}$.

This matches option A exactly. The hypothesis is never the $k+1$ case — that is what the inductive step must prove.

We need the algebraic form of $P(k)$, i.e. the given formula with $n$ replaced by $k$.

Option B states the formula for $n=k+1$, which is what we need to prove in the inductive step, not what we assume.

Option C writes the sum up to $k$ on the left but uses $(k+1)(k+2)(k+3)$ on the right — this is inconsistent, mixing indices from $P(k)$ and $P(k+1)$.

Option D is the formula for the sum of squares $\sum r^2$, which is a completely different identity and irrelevant here.

Option A correctly states $P(k)$: the left side sums $r(r+1)$ up to $k$, and the right side uses the formula evaluated at $k$.