How many VCAA Specialist Mathematics questions cover Discrete mathematics?

AusGrader has 29 VCAA Specialist Mathematics questions on Discrete mathematics, all with instant AI grading and detailed marking feedback.

VCE Specialist Mathematics — Discrete mathematics Questions & Answers

Q3

2021

QCAA

Paper 2

1 mark

Q3

1 mark

Given $n \in Z^+$ , for which proposition can the initial statement for mathematical induction be proven?

A

$x^{2n} - y^{2n}$ is divisible by $(x+y)\forall(x+y) \neq 0$

B

$1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6}n(2n^2 + 3n - 1)$

C

$(n+1)^3 + (n+2)^3$ is divisible by 3

D

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

Reveal Answer

A

$x^{2n} - y^{2n}$ is divisible by $(x+y)\forall(x+y) \neq 0$

Correct Answer

For the base case $n=1$ , the expression becomes $x^2 - y^2 = (x-y)(x+y)$ , which is clearly divisible by $(x+y)$ . Since the initial statement is true, mathematical induction can be applied.

B

$1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6}n(2n^2 + 3n - 1)$

For $n=1$ , the LHS is $1^2 = 1$ , but the RHS is $\frac{1}{6}(1)(2(1)^2 + 3(1) - 1) = \frac{4}{6} = \frac{2}{3}$ . Since $1 \neq \frac{2}{3}$ , the initial statement is false.

C

$(n+1)^3 + (n+2)^3$ is divisible by 3

For $n=1$ , the expression is $(1+1)^3 + (1+2)^3 = 2^3 + 3^3 = 8 + 27 = 35$ . Since 35 is not divisible by 3, the initial statement is false.

D

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

For $n=1$ , the LHS is $\frac{1}{(2(1)-1)(2(1)+1)} = \frac{1}{3}$ , but the RHS is $\frac{1}{1+1} = \frac{1}{2}$ . Since $\frac{1}{3} \neq \frac{1}{2}$ , the initial statement is false.

Q8

2024

VCAA

Paper 1

4 marks

Q8

4 marks

Prove by mathematical induction that
$1 \times 7 + 2 \times 15 + 3 \times 23 + \dots + n(8n - 1) = \frac{1}{6}n(n + 1)(16n + 5)$ for all $n \in N$ .

Reveal Answer

For the case $n=1$ in the proposition, the left-hand side is $1\times7=7$ , and the right side is $\frac{1}{6}\times2\times21=7$ .

Therefore, the proposition is true for $n=1$ .

Assume that the proposition is true for $n=k$ :

$1\times7+2\times15+3\times23+\ldots+k(8k-1)=\frac{1}{6}k(k+1)(16k+5)$

Then the left-hand side with $n=k+1$ is

$1\times7+2\times15+3\times23+\ldots+k(8k-1)+(k+1)(8(k+1)-1)$

$=\frac{1}{6}k(k+1)(16k+5)+(k+1)(8(k+1)-1)$

$=(k+1)\left(\frac{1}{6}k(16k+5)+(8k+7)\right)$

$=\frac{1}{6}(k+1)(16k^2+53k+42)$

$=\frac{1}{6}(k+1)(k+2)(16k+21)$

which is equal to the right-hand side with $n=k+1$ .

Therefore, by the principle of mathematical induction, the proposition is true for all $n\in\mathbb{N}$ .

Marking Criteria

Descriptor	Marks
Proves the base case by showing the proposition is true for $n=1$	1
States the inductive assumption that the proposition is true for $n=k$	1
Correctly substitutes the inductive assumption into the expression for $n=k+1$	1
Completes the algebraic manipulation to prove the $n=k+1$ case and provides a valid concluding statement	1

Q12

2022

QCAA

Paper 1

6 marks

Q12

6 marks

Given $z_1 = a + bi$ , $z_2 = c + di$ $\forall a, b, c, d \in R$ , and $z_2 \neq 0$ , prove the identity
$\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$

Reveal Answer

Prove $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$ , given $z_1 = a+bi, z_2 = c+di, z_2 \neq 0$

LHS $= \left| \frac{a+bi}{c+di} \right|$
$= \left| \frac{(a+bi)(c-di)}{(c+di)(c-di)} \right|$
$= \left| \frac{ac - adi + bci - bdi^2}{c^2 + d^2} \right|$
$= \left| \frac{(ac+bd) + (bc-ad)i}{c^2+d^2} \right|$
$= \sqrt{\frac{(ac+bd)^2 + (bc-ad)^2}{(c^2+d^2)^2}}$
$= \sqrt{\frac{(ac)^2 + 2abcd + (bd)^2 + (bc)^2 - 2abcd + (ad)^2}{(c^2+d^2)^2}}$
$= \sqrt{\frac{(ac)^2 + (bd)^2 + (bc)^2 + (ad)^2}{(c^2+d^2)^2}}$
$= \sqrt{\frac{a^2(c^2+d^2) + b^2(c^2+d^2)}{(c^2+d^2)^2}}$
$= \sqrt{\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}}$
$= \sqrt{\frac{a^2+b^2}{c^2+d^2}}$
$= \frac{|z_1|}{|z_2|}$
= RHS QED

Marking Criteria

Descriptor	Marks
Correctly multiplies the numerator and denominator by the complex conjugate of z₂	1
Realises denominator (in simplest form) and expands numerator	1
Determines modulus of the expression	1
Simplifies numerator	1
Factorises numerator	1
Completes the proof using mathematical reasoning	1

Q16

2020

QCAA

Paper 1

6 marks

Q16

6 marks

Given $\cos(\theta) \neq 0 \forall n \in Z^+$ , use mathematical induction to prove

$\cos(\theta) - \cos(3\theta) + \cos(5\theta) - \dots + (-1)^{n+1} \cos((2n-1)\theta) = \frac{1-(-1)^n \cos(2n\theta)}{2\cos(\theta)}$

Reveal Answer

$\cos(\theta) - \cos(3\theta) + \cos(5\theta) - \dots + (-1)^{n+1} \cos((2n - 1)\theta) = \frac{1-(-1)^n \cos(2n\theta)}{2 \cos(\theta)}$
Let $P(n)$ represent the proposition.

R.T.P. $P(1)$ is true.
LHS $= \cos \theta$
RHS $= \frac{1 - (-1) \cos(2\theta)}{2 \cos \theta}$
$= \frac{1 + \cos(2\theta)}{2 \cos \theta}$
$= \frac{1 + 2 \cos^2 \theta - 1}{2 \cos \theta}$
$= \cos \theta$
$= \text{RHS}$
$\therefore P(1)$ is true.

Assume $P(k)$ is true.
$\cos(\theta) - \cos(3\theta) + \cos(5\theta) - \dots + (-1)^{k+1} \cos((2k - 1)\theta) = \frac{1 - (-1)^k \cos(2k\theta)}{2 \cos(\theta)}$

R.T.P. $P(k + 1)$ is true.
$\cos(\theta) - \cos(3\theta) + \cos(5\theta) - \dots + (-1)^{k+1} \cos((2k - 1)\theta) + (-1)^{(k+1)+1} \cos((2(k + 1) - 1)\theta)$
$= \frac{1 - (-1)^{k+1} \cos(2(k + 1)\theta)}{2 \cos(\theta)}$

LHS $= \frac{1 - (-1)^k \cos(2k\theta)}{2 \cos(\theta)} + (-1)^{k+2} \cos((2k + 1)\theta)$
$= \frac{1 - (-1)^k \cos(2k\theta) + (-1)^{k+2} \cos((2k + 1)\theta) 2\cos(\theta)}{2 \cos(\theta)}$
$= \frac{1 - (-1)^k \cos(2k\theta) + (-1)^{k+2} (\cos[(2k + 1)\theta - \theta] + \cos[(2k + 1)\theta + \theta])}{2 \cos(\theta)}$
$= \frac{1 - (-1)^k \cos(2k\theta) + (-1)^{k+2} \cos(2k\theta) + (-1)^{k+2} \cos((2k + 2)\theta)}{2 \cos(\theta)}$
$= \frac{1 + (-1)^{k+2} \cos((2k + 2)\theta)}{2 \cos(\theta)}$
$= \frac{1 + (-1)^{k+1}(-1)^1 \cos(2(k + 1)\theta)}{2 \cos(\theta)}$
$= \frac{1 - (-1)^{k+1} \cos(2(k + 1)\theta)}{2 \cos(\theta)}$
$= \text{RHS}$
So $P(k + 1)$ is true. By mathematical induction, the formula is true for $n = 1, 2, \dots$

Marking Criteria

Descriptor	Marks
correctly proves the initial statement	1
correctly states the assumption and the proof requirement for the inductive step	1
uses assumption in the proof of the inductive step	1
determines a simplified expression based on the use of a common denominator and a suitable trigonometric product identity	1
determines a simplified expression based on the recognition that $(-1)^k = (-1)^{k+2}$	1
completes proof and communicates a suitable conclusion	1

Q13

2021

QCAA

Paper 1

6 marks

Q13

6 marks

Use $z = a + bi$ and $w = c + di$ , where $a, b, c, d \in R$ , to prove

$|z - w|^2 = |z|^2 + |w|^2 - 2Re(z\bar{w})$

Reveal Answer

RTP
$|z - w|^2 = |z|^2 + |w|^2 - 2\text{Re}(z \overline{w})$

LHS $= |z - w|^2$
$= |(a + bi) - (c + di)|^2$
$= |(a - c) + (b - d)i|^2$
$= (a - c)^2 + (b - d)^2$
$= a^2 - 2ac + c^2 + b^2 - 2bd + d^2$

RHS $= |z|^2 + |w|^2 - 2\text{Re}(z \overline{w})$
$= |a + bi|^2 + |c + di|^2 \dots$
$\dots - 2\text{Re}((a + bi)(c - di))$
$= a^2 + b^2 + c^2 + d^2 \dots$
$\dots - 2\text{Re}((ac + bd) + (bc - ad)i)$
$= a^2 + b^2 + c^2 + d^2 - 2(ac + bd)$
$= a^2 + b^2 + c^2 + d^2 - 2ac - 2bd$
$= a^2 - 2ac + c^2 + b^2 - 2bd + d^2$
$= \text{LHS}$

Marking Criteria

Descriptor	Marks
correctly expresses $z - w$ in terms of $a, b, c$ and $d$ in Cartesian form	1
expresses $\|z - w\|^2$ in terms of $a, b, c$ and $d$ in expanded form	1
correctly expresses $\|z\|^2 + \|w\|^2$ in terms of $a, b, c$ and $d$ in expanded form	1
correctly expresses $(z \overline{w})$ in terms of $a, b, c$ and $d$	1
completes proof	1
shows logical organisation, communicating key steps	1

Q15

2025

QCAA

Paper 2

6 marks

Q15

6 marks

De Moivre's theorem can be expressed as

(r(\cos(\theta) + i\sin(\theta)))^n = r^n(\cos(n\theta) + i\sin(n\theta)) \quad \forall n \in Z^+

Prove De Moivre's theorem using mathematical induction.

Reveal Answer

Initial statement
Prove the rule is true for $n = 1$ .
$\text{LHS} = r(\cos(\theta) + i\sin(\theta))$
$\text{RHS} = r^1(\cos(1 \times \theta) + i\sin(1 \times \theta)) = r(\cos(\theta) + i\sin(\theta)) = \text{LHS}$

Assume the rule is true for $n = k$ .

(r(\cos(\theta) + i\sin(\theta)))^k = r^k(\cos(k\theta) + i\sin(k\theta))

Inductive step
Prove the rule is true for $n = k + 1$ .

(r(\cos(\theta) + i\sin(\theta)))^{k+1} = r^{k+1}(\cos((k+1)\theta) + i\sin((k+1)\theta))

\begin{align*} \text{LHS} &= (r(\cos(\theta) + i\sin(\theta)))^k \cdot r(\cos(\theta) + i\sin(\theta))^1\\ &= r^k(\cos(k\theta) + i\sin(k\theta)) \ r(\cos(\theta) + i\sin(\theta))\\ &= r^{k+1}(\cos(k\theta)\cos(\theta) + i\cos(k\theta)\sin(\theta) + i\sin(k\theta)\cos(\theta) - \sin(k\theta)\sin(\theta))\\ &= r^{k+1}(\cos(k\theta)\cos(\theta) - \sin(k\theta)\sin(\theta) + i(\sin(k\theta)\cos(\theta) + \cos(k\theta)\sin(\theta)))\\ & = r^{k+1}(\cos(k\theta + \theta) + i(\sin(k\theta + \theta)))\\ & = r^{k+1}(\cos((k+1)\theta) + i\sin((k+1)\theta))\\ &= \text{RHS} \end{align*}

Conclusion:
The rule is proven true for $n = k + 1$ . By mathematical induction, the rule is true for $n = 1, 2, \dots$

Marking Criteria

Descriptor	Marks
correctly proves the initial statement	1
correctly states a suitable assumption	1
uses the assumption statement	1
expresses result in Cartesian form	1
uses angle sum and difference identities	1
completes proof and states a suitable conclusion	1

Q2

2023

QCAA

Paper 1

1 mark

Q2

1 mark

Consider the proof of the following proposition using mathematical induction.

$\sum_{r=1}^n r(r+1) = \frac{1}{3}n(n+1)(n+2) \forall n \in Z^+$

An appropriate assumption statement within the proof is

A

$\sum_{r=1}^k k(k+1) = \frac{1}{3}k(k+1)(k+2)$

B

$\sum_{r=1}^k k(k+1) = \frac{1}{3}n(n+1)(n+2)$

C

$\sum_{r=1}^k r(r+1) = \frac{1}{3}k(k+1)(k+2)$

D

$\sum_{r=1}^k r(r+1) = \frac{1}{3}n(n+1)(n+2)$

Reveal Answer

A

$\sum_{r=1}^k k(k+1) = \frac{1}{3}k(k+1)(k+2)$

This option incorrectly replaces the summation variable $r$ with the limit $k$ inside the sum. The term must remain $r(r+1)$ to represent the sum of the series, whereas $\sum k(k+1)$ would imply summing a constant term $k$ times.

B

$\sum_{r=1}^k k(k+1) = \frac{1}{3}n(n+1)(n+2)$

This option makes two errors: it incorrectly uses $k$ inside the summation instead of the index variable $r$ , and it fails to replace $n$ with $k$ on the right-hand side of the equation.

C

$\sum_{r=1}^k r(r+1) = \frac{1}{3}k(k+1)(k+2)$

Correct Answer

This is the correct inductive hypothesis (assumption step). It assumes the statement is true for $n=k$ by replacing every instance of $n$ in the original proposition with $k$ .

D

$\sum_{r=1}^k r(r+1) = \frac{1}{3}n(n+1)(n+2)$

This option incorrectly mixes variables. While the summation limit is changed to $k$ , the right-hand side still uses $n$ . The inductive hypothesis requires substituting $k$ for $n$ on both sides of the equation.

Q1

2024

VCAA

Paper 2

1 mark

Q1

1 mark

Consider the following proof.

Prove that $\sqrt{15} + \sqrt{7} > \sqrt{19}$ .

Assume $\sqrt{15} + \sqrt{7} \le \sqrt{19}$ .

Then $\left(\sqrt{15} + \sqrt{7}\right)^2 \le 19$

$15 + 2\sqrt{105} + 7 \le 19$

$2\sqrt{105} \le -3$

hence $\sqrt{15} + \sqrt{7} > \sqrt{19}$ .

This proof can be best described as a

A

direct proof.

B

proof by contrapositive.

C

proof by contradiction.

D

proof by counter-example.

E

proof by mathematical induction.

Reveal Answer

A

direct proof.

A direct proof would start with known facts and logically deduce the conclusion step-by-step, rather than assuming the opposite of what needs to be proven.

B

proof by contrapositive.

A proof by contrapositive is used for conditional statements ( $P \implies Q$ ) by proving that $\neg Q \implies \neg P$ . This proof does not involve a conditional statement.

C

proof by contradiction.

Correct Answer

This is a proof by contradiction because it assumes the negation of the statement to be proved ( $\sqrt{15} + \sqrt{7} \le \sqrt{19}$ ) and derives a mathematical impossibility ( $2\sqrt{105} \le -3$ ), thereby proving the original statement must be true.

D

proof by counter-example.

A proof by counter-example is used to disprove a universal statement by providing a specific case where it fails, which is not what is happening here.

E

proof by mathematical induction.

Mathematical induction is used to prove statements for all natural numbers by establishing a base case and an inductive step, which does not apply to proving a single inequality.

Q15

2023

QCAA

Paper 1

5 marks

Q15

The sum of a geometric progression with $n$ terms, where the first term is 1 and the common ratio is $r$ , is given by

$1 + r + r^2 + r^3 + ... + r^{n-1} = \frac{r^n - 1}{r - 1} \quad (\text{for } r \neq 1).$

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

Q15a

1 mark

Initial statement:

Reveal Answer

Initial statement
Prove the rule is true for $n=1$ .
$LHS = 1$
$RHS = \frac{r-1}{r-1}$
$= 1 = LHS$

Marking Criteria

Descriptor	Marks
Correctly proves the initial statement	1

Q15b

3 marks

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

$1 + r + r^2 + r^3 + ... + r^{k-1} = \frac{r^k - 1}{r - 1} \quad (r \neq 1).$

Inductive step:

Reveal Answer

Given assumption
$1 + r + r^2 + r^3 + ... + r^{k-1} = \frac{r^k-1}{r-1} \quad (r \neq 1)$

Inductive step
Prove the rule is true for $n=k+1$ for $r \neq 1$
$1 + r + r^2 + r^3 + ... + r^{k-1} + r^k = \frac{r^{k+1}-1}{r-1}$

$LHS = \frac{r^k-1}{r-1} + r^k$
$= \frac{r^k-1 + r^{k+1} - r^k}{r-1}$
$= \frac{r^{k+1}-1}{r-1}$
$= RHS$

Marking Criteria

Descriptor	Marks
Correctly establishes an expression representing the left-hand side requirement of the inductive step proof	1
Uses the given assumption within the inductive step proof	1
Shows mathematical reasoning to complete the inductive step proof	1

Q15c

1 mark

Conclusion:

Reveal Answer

Conclusion
The rule is proven true for $n=k+1$ .
By mathematical induction, the rule is true for $n=1, 2, ...$

Marking Criteria

Descriptor	Marks
States a suitable conclusion to the proof	1

Q2

2024

VCAA

Paper 1

3 marks

Q2

3 marks

Prove that if $x$ is an odd integer then $2x^2 - 3x - 7$ is even, using a direct proof.

Reveal Answer

As $x$ is an odd integer, we may let $x = 2k + 1$ where $k \in Z$ .

\begin{align*} 2x^2 - 3x - 7 &= 2(2k + 1)^2 - 3(2k + 1) - 7\\ &= 2(4k^2 + 4k + 1) - 6k - 3 - 7\\ &= 8k^2 + 2k - 8\\ &= 2(4k^2 + k - 4) \text{ which is even} \end{align*}

Marking Criteria

Descriptor	Marks
Provides a complete and correct proof using algebraic substitution, demonstrating that the expression factors to a multiple of 2	3
Correctly substitutes $x = 2k + 1$ and expands the expression, but provides an incomplete proof or makes a minor algebraic error	2
Substitutes $x = 2k + 1$ (or equivalent) into the expression	1
None of the above	0

Q18

2022

QCAA

Paper 1

5 marks

Q18

5 marks

It is proposed that the following expression is divisible by $(1 + \text{cis}(\theta))$ for $n \in Z^+$ , $(1 + \text{cis}(\theta)) \neq 0$ .

$\sum_{r=0}^{2n+1} \text{cis}(r\theta)$

Evaluate the reasonableness of the proposition.

Reveal Answer

Mathematical induction can be used to prove the proposition
$\sum_{r=0}^{2n+1} \text{cis}(r\theta)$ is divisible by $(1 + \text{cis}(\theta))$ for $n \in \mathbb{Z}^+$
Let $n = 1$
$\sum_{r=0}^{3} \text{cis}(r\theta) = \text{cis}(0) + \text{cis}(\theta) + \text{cis}(2\theta) + \text{cis}(3\theta)$
$= 1 + \text{cis}(\theta) + (\text{cis}(\theta))^2 + (\text{cis}(\theta))^3$
$= (1 + \text{cis}(\theta))(1 + (\text{cis}(\theta))^2)$
This expression is divisible by $(1 + \text{cis}(\theta))$ , so the proposition is true for $n = 1$ .

Assume $n = k$ is true for $k \in \mathbb{Z}^+$ :
$\sum_{r=0}^{2k+1} \text{cis}(r\theta) = (1 + \text{cis}(\theta))Q(\theta)$ where $Q(\theta)$ is a function of $\theta$

Let $n = k + 1$
$\sum_{r=0}^{2k+3} \text{cis}(r\theta) = \sum_{r=0}^{2k+1} \text{cis}(r\theta) + \text{cis}((2k+2)\theta) + \text{cis}((2k+3)\theta)$
$= (1 + \text{cis}(\theta))Q(\theta) + (\text{cis}(\theta))^{2k+2} + (\text{cis}(\theta))^{2k+3}$
$= (1 + \text{cis}(\theta))Q(\theta) + (\text{cis}(\theta))^{2k+2}(1 + \text{cis}(\theta))$
$= (1 + \text{cis}(\theta))(Q(\theta) + (\text{cis}(\theta))^{2k+2})$
$= (1 + \text{cis}(\theta))R(\theta)$ where $R(\theta)$ is a function of $\theta$
The proposition is true for $n = k + 1$ . By mathematical induction, the formula is true for $n = 1, 2, ...$

Marking Criteria

Descriptor	Marks
correctly proves the initial statement	1
correctly establishes an appropriate assumption for $n=k$	1
expresses the sum based on $n=k+1$ in terms of the assumption	1
expresses a result using a common factor of $(1 + \text{cis}(\theta))$	1
proves the inductive step	1

Q8

2021

QCAA

Paper 1

1 mark

Q8

1 mark

Let $P(n)$ be the proposition that

$\sum_{r=1}^n (r+1)3^{r-1} = n \times 3^n \forall n \in Z^+$

Which option represents a correct formulation of the assumption that $P(k)$ is true $\forall k \in Z^+$ in a proof using mathematical induction?

A

$\sum_{r=1}^k (k+1)3^{k-1} = k \times 3^k$

B

$\sum_{r=1}^k (k+1)3^{k-1} = n \times 3^n$

C

$\sum_{r=1}^k (r+1)3^{r-1} = k \times 3^k$

D

$\sum_{r=1}^k (r+1)3^{r-1} = r \times 3^r$

Reveal Answer

A

$\sum_{r=1}^k (k+1)3^{k-1} = k \times 3^k$

This option incorrectly replaces the summation index $r$ with the upper limit $k$ inside the sigma notation, which would make the term constant rather than varying with the index.

B

$\sum_{r=1}^k (k+1)3^{k-1} = n \times 3^n$

This option incorrectly replaces $r$ with $k$ inside the sum and fails to replace $n$ with $k$ on the right-hand side of the equation.

C

$\sum_{r=1}^k (r+1)3^{r-1} = k \times 3^k$

Correct Answer

To state the assumption $P(k)$ , one substitutes $n$ with $k$ in the original proposition: the upper limit of the sum becomes $k$ and the right-hand side becomes $k \times 3^k$ .

D

$\sum_{r=1}^k (r+1)3^{r-1} = r \times 3^r$

The right-hand side of the equation represents the total value of the sum and must depend on the upper limit $k$ , not the dummy index variable $r$ .

Q8

2023

VCAA

Paper 1

4 marks

Q8

4 marks

A function $f$ has the rule $f(x) = x e^{2x}$ .

Use mathematical induction to prove that $f^{(n)}(x) = (2^n x + n 2^{n-1}) e^{2x}$ for $n \in Z^+$ , where $f^{(n)}(x)$ represents the $n^{\text{th}}$ derivative of $f(x)$ . That is, $f(x)$ has been differentiated $n$ times.

Reveal Answer

\begin{align*} f'(x) &= e^{2x} + 2xe^{2x} \\ &= (2x+1)e^{2x} \\ & = (2^1x + 1 \times 2^{1-1})e^{2x} \end{align*}

and so the statement is true for $n=1$ .

Assume that the statement is true for some $k > 1$ . That is

f^{(k)}(x) = (2^kx + k2^{k-1})e^{2x}

then

\begin{align*} f^{(k+1)}(x) &= \frac{d}{dx}\left((2^kx + k2^{k-1})e^{2x}\right) \\ &= 2^ke^{2x} + 2(2^kx + k2^{k-1})e^{2x} \\ & = (2^{k+1}x + (k+1)2^k)e^{2x} \end{align*}

and so the statement is true for $n = k+1$ .
Therefore, by the principle of mathematical induction, the statement is true for all $n \in Z^+$ .

Marking Criteria

Descriptor	Marks
Shows the statement is true for $n=1$ by correctly finding $f'(x)$ and showing it matches the formula	1
States the inductive assumption that the statement is true for $n=k$	1
Correctly differentiates $f^{(k)}(x)$ with respect to $x$ to find $f^{(k+1)}(x)$	1
Simplifies the expression to show it matches the formula for $n=k+1$ and provides a valid concluding statement	1

Q3

2024

QCAA

Paper 1

1 mark

Q3

1 mark

Consider a proof of the proposition $\sum_{j=1}^{n} (2j-1) = n^2 \ \forall n \in Z^+$ using mathematical induction.

Within the proof of the inductive step, the proposition for $n=k+1$ could be expressed as

A

$\sum_{j=1}^{k+1} (2j-1) = k^2 + 2k + 1$

B

$\sum_{j=1}^{k+1} (2k+1) = k^2 + 2k + 1$

C

$\sum_{j=1}^{k+1} (2j-1) = k^2 + 1$

D

$\sum_{j=1}^{k+1} (2k+1) = k^2 + 1$

Reveal Answer

A

$\sum_{j=1}^{k+1} (2j-1) = k^2 + 2k + 1$

Correct Answer

To express the proposition for $n=k+1$ , substitute $k+1$ for $n$ on both sides of the equation. The upper limit of the sum becomes $k+1$ , and the right side becomes $(k+1)^2$ , which expands to $k^2 + 2k + 1$ .

B

$\sum_{j=1}^{k+1} (2k+1) = k^2 + 2k + 1$

The term inside the summation must remain a function of the index variable $j$ (specifically $2j-1$ ). Replacing it with $2k+1$ incorrectly makes the term constant with respect to the summation index.

C

$\sum_{j=1}^{k+1} (2j-1) = k^2 + 1$

The right side of the equation represents $n^2$ evaluated at $n=k+1$ . This should be $(k+1)^2 = k^2 + 2k + 1$ , not $k^2 + 1$ .

D

$\sum_{j=1}^{k+1} (2k+1) = k^2 + 1$

This option is incorrect on both sides: the summation term should remain $(2j-1)$ , and the right side should be the expansion of $(k+1)^2$ , which is $k^2 + 2k + 1$ .

Q2

2020

QCAA

Paper 1

1 mark

Q2

1 mark

When using proof by mathematical induction to show that $n(2n-1)(2n+1)$ is divisible by $3 \forall n \in Z^+$ , the inductive step requires proving

A

$(k+1)(2k)(2k+2)$ is divisible by 3.

B

$(k+1)(2k)(2k+3)$ is divisible by 3.

C

$(k+1)(2k+1)(2k+2)$ is divisible by 3.

D

$(k+1)(2k+1)(2k+3)$ is divisible by 3.

Reveal Answer

A

$(k+1)(2k)(2k+2)$ is divisible by 3.

This option incorrectly calculates the terms involving $n$ . Substituting $n=k+1$ into $(2n-1)$ yields $2k+1$ , not $2k$ , and into $(2n+1)$ yields $2k+3$ , not $2k+2$ .

B

$(k+1)(2k)(2k+3)$ is divisible by 3.

The middle term is incorrect. When substituting $n=k+1$ into $(2n-1)$ , the result is $2(k+1)-1 = 2k+1$ , not $2k$ .

C

$(k+1)(2k+1)(2k+2)$ is divisible by 3.

The final term is incorrect. When substituting $n=k+1$ into $(2n+1)$ , the result is $2(k+1)+1 = 2k+3$ , not $2k+2$ .

D

$(k+1)(2k+1)(2k+3)$ is divisible by 3.

Correct Answer

In the inductive step, we replace $n$ with $k+1$ . The expression $n(2n-1)(2n+1)$ becomes $(k+1)(2(k+1)-1)(2(k+1)+1)$ , which simplifies to $(k+1)(2k+1)(2k+3)$ .

VCAA Specialist Mathematics Discrete mathematics

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