Question 8
Let $P(n)$ be the proposition that
$$\sum_{r=1}^{n}(r+1)3^{r-1}=n\times3^n\quad\forall n\in\mathbb{Z}^+$$
Which option represents a correct formulation of the assumption that $P(k)$ is true $\forall k\in\mathbb{Z}^+$ in a proof using mathematical induction?
Not quite. The counter $r$ has been replaced by $k$ as well. Then every term in the sum is the same, $(k+1)3^{k-1}$, and the left side adds up to $k(k+1)3^{k-1}$. Only $n$ changes to $k$; $r$ stays as the counter. In 2021, 34.2% of Queensland students chose (A).
Not quite. There are two slips here. The counter $r$ has been replaced by $k$, so every term in the sum is the same, and the right-hand side still has $n$. In $P(k)$ every $n$ becomes $k$ and $r$ stays inside the sum. In 2021, 4.5% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2021, 59.9% of Queensland students chose (C).
Not quite. The left side is right, but the right-hand side uses $r$. $r$ is the counter that runs from $1$ to $k$ inside the sum, so it has no single value outside it. Replacing $n$ with $k$ gives $k\times3^k$. In 2021, 1.3% of Queensland students chose (D).
Build the assumption piece by piece. The panel writes out the sum for a small value of $k$, so you can see what each choice does.
This is $P(k)$. Only the top of the sum and the right-hand side changed. The counter $r$ still runs $1,2,\ldots,k$, so each term is different.
With $k$ inside the sum, every term is the same, $(k+1)3^{k-1}$. Adding $k$ copies gives $k(k+1)3^{k-1}$, which is a different statement. Keep $r$ as the counter.
The top of the sum is still $n$. In $P(k)$ every $n$ becomes $k$, including the top of the sum.
The right-hand side still has $n$. Replace every $n$ with $k$.
$r$ is the counter inside the sum. Outside the sum it has no single value, so $r\times3^r$ can’t be the right-hand side.
$P(k)$ is $P(n)$ with $n$ replaced by $k$. Find every $n$ in the statement. There are two: the top of the sum and the right-hand side.
$r$ is a counter. It runs through $1,2,\ldots,k$ and is used up inside the sum, so it stays as $r$. For example, with $k=3$,
$$\begin{aligned}&\sum_{r=1}^{3}(r+1)3^{r-1}\\&\quad=(1+1)3^0+(2+1)3^1\\&\qquad+(3+1)3^2\end{aligned}$$
Replace both copies of $n$ with $k$ and leave $r$ alone.
$$\sum_{r=1}^{k}(r+1)3^{r-1}=k\times3^k$$
Replace every $n$ with $k$. The counter $r$ stays inside the sum.
$$\sum_{r=1}^{k}(r+1)3^{r-1}=k\times3^k$$
One thing I noticed: as printed, $P(1)$ reads $2=3$, so this proposition is false and the proof could never be finished. With $(2r+1)$ in place of $(r+1)$ it is true. The question only tests how you write the assumption, so the answer is (C) either way.
How Queensland students answered in 2021
From the QCAA Specialist Mathematics subject report for 2021. 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 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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