Specialist Mathematics · Mock exam 2026 All mock exam questions
Mock exam · Paper 1 · Simple familiar

Question 12

[5 marks] Technology-free
The question

Consider the proposition

$$\sum_{r=1}^{n}r\,2^{r}=(n-1)\,2^{n+1}+2$$

Prove that this proposition is true $\forall\, n\in Z^{+}$ using mathematical induction by completing the steps of the proof as indicated.

a)

Initial statement:

[1 mark]

Assuming the proposition is true for $n=k$,

$$\sum_{r=1}^{k}r\,2^{r}=(k-1)\,2^{k+1}+2.$$

b)

Inductive step:

[3 marks] c)

Conclusion:

[1 mark]
Watch the situation first

Each sum is the one before plus one more term

The formula predicts every total. Watch each new term carry the bar exactly onto the next prediction, then see how one case hands on to the next.

The formula (n − 1) 2n+1 + 2 predicts each sumStart at n = 1: one term, 1 × 21 = 2. It matchesAssume n = k works, then add the (k + 1)th termThe new total lands exactly on k 2k+2 + 2Each case hands the result on to the next oneTrue for n = 1, and k ⇒ k + 1, so true for all n ∈ Z+sum of r 2r, to scale(n − 1) 2n+1 + 2n = 1n = 2n = 3n = 42103498⇓⇓⇓⋮LHS = 1 × 21 = 2     RHS = (1 − 1) 22 + 2 = 2(k − 1) 2k+1 + 2 + (k + 1) 2k+1 = 2k · 2k+1 + 2 = k 2k+2 + 2k = 1:   2 + 2 × 22 = 10 = 1 × 23 + 2k = 2:   10 + 3 × 23 = 34 = 2 × 24 + 2k = 3:   34 + 4 × 24 = 98 = 3 × 25 + 2n = 1 ⇒ n = 2 ⇒ n = 3 ⇒ n = 4 ⇒ ⋯

Drawn to scale: one unit of the sum is the same length in every row. Blue is the sum up to $k$, which the assumption gives you. Orange is the $(k+1)$th term.

Now for the mathematics

Add the right term and aim at the right target

The blue bar is the sum up to $k$, which the assumption gives you as $(k-1)\,2^{k+1}+2$. Pick the term you add on the end and the target you aim for. The bar should land exactly on the target, for every $k$.

target sum to k term 0 0 LHS
Sum up to $k$, assumed The term you add Drawn to scale for each $k$
The (k + 1)th term you add
The target you aim for

The bar lands exactly on the target. That is the inductive step:

$$\begin{aligned}&(k-1)\,2^{k+1}+2+(k+1)\,2^{k+1}\\&\quad=k\,2^{k+2}+2\end{aligned}$$

Move $k$ and it lands every time.

The bar meets this target, but only by luck at this value of $k$. Both choices are slips. Move $k$ and they come apart.

The bar stops short of the target by $2^{k+1}$. You changed only the exponent. The $(k+1)$th term is the general term $r\,2^{r}$ with $r=k+1$, which is $(k+1)\,2^{k+1}$.

The bar stops well short of the target. $k\,2^{k}$ changes nothing: it is the general term $r\,2^{r}$ with $r=k$. Put $r=k+1$ into it instead.

The bar overshoots the target. This target replaced $n$ with $k+1$ in the bracket but with $k$ in the exponent. Replace every $n$ with $k+1$: the target is $k\,2^{k+2}+2$.

Two slips at once. Fix the term first: the $(k+1)$th term is $(k+1)\,2^{k+1}$. Then fix the target: replace every $n$ with $k+1$.

Marking guide · 5 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Choose a method

Step 1 · part a) · Initial statement1 mark

Substitute $n=1$ and evaluate each side separately. The sum has one term, the $r=1$ term.

$$\begin{aligned}\text{LHS}&=1\times2^{1}=2\\\text{RHS}&=(1-1)\,2^{2}+2=2=\text{LHS}\end{aligned}$$

The proposition is true for $n=1$.

Marker: correctly proves the initial statement

Step 2 · part b) · Inductive step1 mark

Assume the proposition is true for $n=k$, then write what you must prove for $n=k+1$.

$$\begin{aligned}\text{Assume:}\quad&\textstyle\sum_{r=1}^{k}r\,2^{r}=(k-1)\,2^{k+1}+2\\\text{Prove:}\quad&\textstyle\sum_{r=1}^{k+1}r\,2^{r}=k\,2^{k+2}+2\end{aligned}$$

Why?

Replace every $n$ with $k+1$: the bracket becomes $(k+1)-1=k$ and the exponent becomes $(k+1)+1=k+2$. The right-hand side is your target.

Marker: correctly establishes an expression representing the left-hand side requirement of the inductive step proof

Step 3 · part b) · Use the assumption1 mark

Split off the $(k+1)$th term, $(k+1)\,2^{k+1}$, then replace the sum to $k$ with the assumption.

$$\begin{aligned}\text{LHS}&=\textstyle\sum_{r=1}^{k}r\,2^{r}+(k+1)\,2^{k+1}\\&=(k-1)\,2^{k+1}+2+(k+1)\,2^{k+1}\end{aligned}$$

Marker: uses the given assumption within the inductive step proof

Step 4 · part b) · Reach the target1 mark

Collect the $2^{k+1}$ terms, using $(k-1)+(k+1)=2k$, then use the index law $2\cdot2^{k+1}=2^{k+2}$.

$$\begin{aligned}\text{LHS}&=2k\cdot2^{k+1}+2\\&=k\,2^{k+2}+2\\&=\text{RHS}\end{aligned}$$

Marker: shows mathematical reasoning to complete the inductive step proof

Step 5 · part c) · Conclusion1 mark

Cite both parts a) and b), then conclude by mathematical induction.

The proposition is true for $n=1$, and if it is true for $n=k$ then it is true for $n=k+1$. By mathematical induction, the proposition is true for $n=1,2,3,\ldots$

Marker: states a suitable conclusion to the proof

Putting it all together

Each phrase of the question gave you something

“$\sum_{r=1}^{n}r\,2^{r}=(n-1)\,2^{n+1}+2$”
The general term is $r\,2^{r}$, so the $(k+1)$th term is $(k+1)\,2^{k+1}$. Replacing every $n$ with $k+1$ gives the target $k\,2^{k+2}+2$.
“true $\forall\, n\in Z^{+}$”
The smallest positive integer is $n=1$, so the initial statement checks $n=1$: LHS $=2$ and RHS $=2$.
“Assuming the proposition is true for $n=k$”
You may replace $\sum_{r=1}^{k}r\,2^{r}$ with $(k-1)\,2^{k+1}+2$ once you have split off the $(k+1)$th term.
“Inductive step”
Write the target for $n=k+1$, split off the last term, use the assumption, collect $2^{k+1}$, then use $2\cdot2^{k+1}=2^{k+2}$.
“Conclusion”
Link both pieces: true for $n=1$, and true for $n=k$ gives true for $n=k+1$. By mathematical induction, true for all $n\in Z^{+}$.
Common errors

If I had to guess where you might lose marks…

  • Wrong $(k+1)$th term. If you add $k\,2^{k+1}$ (changing only the exponent), you get $(2k-1)\,2^{k+1}+2$, which is $2^{k+1}$ short of the target. At $k=1$ it gives $6$ instead of $10$. Adding $k\,2^{k}$ (changing nothing) gives $4$ at $k=1$.
  • Half-updated target. If you replace $n$ with $k+1$ in the bracket but with $k$ in the exponent, your target is $\big((k+1)-1\big)\,2^{k+1}+2$ $=k\,2^{k+1}+2$, which is $6$ at $k=1$ instead of $10$. Correct working then “fails” to reach it, and you either abandon the proof or fudge a line.
  • Index-law slip. Writing $2\cdot2^{k+1}=4^{k+1}$ gives $k\,4^{k+1}+2$, which is $18$ at $k=1$ instead of $10$. At $k=1$, $2\cdot2^{2}=8$ but $4^{2}=16$.
  • Base case at $n=0$. At $n=0$ the empty sum is $0$ and the RHS is $(0-1)\,2^{1}+2=0$, so the check “works”, but $0\notin Z^{+}$. The proof must start at $n=1$.
  • Weak conclusion. Writing only “true for $n=k+1$” or “true for all $n$” without citing both the base case and the inductive step does not explain why induction applies.

Only mark this done when you could do it without help. Reading the solution does not count.

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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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