Question 2
Consider a proof of the proposition $\displaystyle{\sum_{r=1}^{n}r(r+2)}={\frac{n(n+1)(2n+7)}{6}}\ \forall n\in \mathit{Z}^{+}$ using mathematical induction.
Assuming the proposition is true for $n=k$,
$$\sum_{r=1}^{k}r(r+2)=\frac{k(k+1)(2k+7)}{6}.$$
Within the proof of the inductive step, $\displaystyle{\sum_{r=1}^{k+1}r(r+2)}$ could be expressed as
That’s it. Step through the hints below to see why.
Not quite. You added the $k$th term $k(k+2)$ instead of the $(k+1)$th term $(k+1)(k+3)$.
Not quite. You added the formula for the sum of the first $k+1$ terms instead of the $(k+1)$th term. That confuses a sum with a single term.
Not quite. You started from the formula for $n=k+1$, which is the statement you are still trying to prove (circular reasoning). Then you added the $(k+1)$th term on top of a sum that already contains it.
Split off the last term of the sum. A sum to $k+1$ is the sum to $k$ plus one more term, the $(k+1)$th.
$$\begin{aligned}&\sum_{r=1}^{k+1}r(r+2)\\&=\sum_{r=1}^{k}r(r+2)+(k+1)\big((k+1)+2\big)\end{aligned}$$
The inductive step must use the assumption, which is about the sum up to $k$.
Simplify the $(k+1)$th term: substitute $r=k+1$ into $r(r+2)$.
$$(k+1)(k+1+2)=(k+1)(k+3)$$
The general term is $r(r+2)$. The new term is that expression evaluated at $r=k+1$.
Replace the sum to $k$ using the assumption.
$$\begin{aligned}&\sum_{r=1}^{k+1}r(r+2)\\&=\frac{k(k+1)(2k+7)}{6}+(k+1)(k+3)\end{aligned}$$
The inductive hypothesis lets you swap ${\sum_{r=1}^{k}r(r+2)}$ for $\dfrac{k(k+1)(2k+7)}{6}$. This is exactly option (A).
Split off the $(k+1)$th term, which is $(k+1)(k+3)$, then replace the sum to $k$ with the assumption.
$$\begin{aligned}&\sum_{r=1}^{k+1}r(r+2)\\&=\frac{k(k+1)(2k+7)}{6}+(k+1)(k+3)\end{aligned}$$
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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