Question 2
Consider the proof of the following proposition using mathematical induction.
$$\sum_{r=1}^{n}r(r+1)=\frac{1}{3}n(n+1)(n+2)\quad\forall\,n\in\mathbb{Z}^+$$
An appropriate assumption statement within the proof is
Not quite. The summand has to stay as $r(r+1)$. $r$ counts through the terms of the sum. Only the top limit $n$ becomes $k$. In 2023, 15.8% of Queensland students chose (A).
Not quite. The summand changed to $k(k+1)$, and the right-hand side still uses $n$. In 2023, 1.7% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2023, 81.7% of Queensland students chose (C).
Not quite. The right-hand side still has $n$ in it. In the assumption, every $n$ becomes $k$, on both sides. In 2023, 0.7% of Queensland students chose (D).
The right-hand side still depends on $n$, so it can’t be checked for this value of $k$. The assumption must be a statement about $k$ alone.
Both sides match for this value of $k$. This is the proposition written for $n=k$, which is option (C).
They happen to match when $k=1$, but try a larger value. The summand should still be $r(r+1)$.
The two sides don’t match. Changing the summand to $k(k+1)$ makes every term the same, which is a different sum.
The assumption says the rule is true for $n=k$. So you replace every $n$ in the proposition with $k$.
The $r$ in $r(r+1)$ is the counter in the sum. It runs from $1$ up to the top limit, so it stays as $r$.
Only the top limit and the right-hand side change.
$$\sum_{r=1}^{k}r(r+1)=\frac13k(k+1)(k+2)$$
The assumption is the proposition with $n$ replaced by $k$. The counter $r$ is unchanged.
$$\sum_{r=1}^{k}r(r+1)=\frac13k(k+1)(k+2)$$
How Queensland students answered in 2023
From the QCAA Specialist Mathematics subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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