VCAA Specialist Mathematics Discrete mathematics

15 sample questions with marking guides and sample answers

Q3
2021
QCAA
Paper 2
1 mark
Q3
1 mark

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

A

x2n−y2nx^{2n} - y^{2n} is divisible by (x+y)∀(x+y)≠0(x+y)\forall(x+y) \neq 0

B

12+22+32+⋯+n2=16n(2n2+3n−1)1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6}n(2n^2 + 3n - 1)

C

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

D

∑r=1n1(2r−1)(2r+1)=nn+1\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)} = \frac{n}{n+1}

Reveal Answer
A

x2n−y2nx^{2n} - y^{2n} is divisible by (x+y)∀(x+y)≠0(x+y)\forall(x+y) \neq 0

Correct Answer

For the base case n=1n=1, the expression becomes x2−y2=(x−y)(x+y)x^2 - y^2 = (x-y)(x+y), which is clearly divisible by (x+y)(x+y). Since the initial statement is true, mathematical induction can be applied.

B

12+22+32+⋯+n2=16n(2n2+3n−1)1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6}n(2n^2 + 3n - 1)

For n=1n=1, the LHS is 12=11^2 = 1, but the RHS is 16(1)(2(1)2+3(1)−1)=46=23\frac{1}{6}(1)(2(1)^2 + 3(1) - 1) = \frac{4}{6} = \frac{2}{3}. Since 1≠231 \neq \frac{2}{3}, the initial statement is false.

C

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

For n=1n=1, the expression is (1+1)3+(1+2)3=23+33=8+27=35(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

∑r=1n1(2r−1)(2r+1)=nn+1\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)} = \frac{n}{n+1}

For n=1n=1, the LHS is 1(2(1)−1)(2(1)+1)=13\frac{1}{(2(1)-1)(2(1)+1)} = \frac{1}{3}, but the RHS is 11+1=12\frac{1}{1+1} = \frac{1}{2}. Since 13≠12\frac{1}{3} \neq \frac{1}{2}, the initial statement is false.

Q12
2022
QCAA
Paper 1
6 marks
Q12
6 marks

Given z1=a+biz_1 = a + bi, z2=c+diz_2 = c + di ∀a,b,c,d∈R\forall a, b, c, d \in R, and z2≠0z_2 \neq 0, prove the identity
∣z1z2∣=∣z1∣∣z2∣\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}

Reveal Answer

Prove ∣z1z2∣=∣z1∣∣z2∣\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}, given z1=a+bi,z2=c+di,z2≠0z_1 = a+bi, z_2 = c+di, z_2 \neq 0

LHS =∣a+bic+di∣= \left| \frac{a+bi}{c+di} \right|
=∣(a+bi)(c−di)(c+di)(c−di)∣= \left| \frac{(a+bi)(c-di)}{(c+di)(c-di)} \right|
=∣ac−adi+bci−bdi2c2+d2∣= \left| \frac{ac - adi + bci - bdi^2}{c^2 + d^2} \right|
=∣(ac+bd)+(bc−ad)ic2+d2∣= \left| \frac{(ac+bd) + (bc-ad)i}{c^2+d^2} \right|
=(ac+bd)2+(bc−ad)2(c2+d2)2= \sqrt{\frac{(ac+bd)^2 + (bc-ad)^2}{(c^2+d^2)^2}}
=(ac)2+2abcd+(bd)2+(bc)2−2abcd+(ad)2(c2+d2)2= \sqrt{\frac{(ac)^2 + 2abcd + (bd)^2 + (bc)^2 - 2abcd + (ad)^2}{(c^2+d^2)^2}}
=(ac)2+(bd)2+(bc)2+(ad)2(c2+d2)2= \sqrt{\frac{(ac)^2 + (bd)^2 + (bc)^2 + (ad)^2}{(c^2+d^2)^2}}
=a2(c2+d2)+b2(c2+d2)(c2+d2)2= \sqrt{\frac{a^2(c^2+d^2) + b^2(c^2+d^2)}{(c^2+d^2)^2}}
=(a2+b2)(c2+d2)(c2+d2)2= \sqrt{\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}}
=a2+b2c2+d2= \sqrt{\frac{a^2+b^2}{c^2+d^2}}
=∣z1∣∣z2∣= \frac{|z_1|}{|z_2|}
= RHS QED

Marking Criteria
DescriptorMarks

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⁡(θ)≠0∀n∈Z+\cos(\theta) \neq 0 \forall n \in Z^+, use mathematical induction to prove

cos⁡(θ)−cos⁡(3θ)+cos⁡(5θ)−⋯+(−1)n+1cos⁡((2n−1)θ)=1−(−1)ncos⁡(2nθ)2cos⁡(θ)\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⁡(θ)−cos⁡(3θ)+cos⁡(5θ)−⋯+(−1)n+1cos⁡((2n−1)θ)=1−(−1)ncos⁡(2nθ)2cos⁡(θ)\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)P(n) represent the proposition.

R.T.P. P(1)P(1) is true.
LHS =cos⁡θ= \cos \theta
RHS =1−(−1)cos⁡(2θ)2cos⁡θ= \frac{1 - (-1) \cos(2\theta)}{2 \cos \theta}
=1+cos⁡(2θ)2cos⁡θ= \frac{1 + \cos(2\theta)}{2 \cos \theta}
=1+2cos⁡2θ−12cos⁡θ= \frac{1 + 2 \cos^2 \theta - 1}{2 \cos \theta}
=cos⁡θ= \cos \theta
=RHS= \text{RHS}
∴P(1)\therefore P(1) is true.

Assume P(k)P(k) is true.
cos⁡(θ)−cos⁡(3θ)+cos⁡(5θ)−⋯+(−1)k+1cos⁡((2k−1)θ)=1−(−1)kcos⁡(2kθ)2cos⁡(θ)\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)P(k + 1) is true.
cos⁡(θ)−cos⁡(3θ)+cos⁡(5θ)−⋯+(−1)k+1cos⁡((2k−1)θ)+(−1)(k+1)+1cos⁡((2(k+1)−1)θ)\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)
=1−(−1)k+1cos⁡(2(k+1)θ)2cos⁡(θ)= \frac{1 - (-1)^{k+1} \cos(2(k + 1)\theta)}{2 \cos(\theta)}

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

Marking Criteria
DescriptorMarks

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)^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+biz = a + bi and w=c+diw = c + di, where a,b,c,d∈Ra, b, c, d \in R, to prove

∣z−w∣2=∣z∣2+∣w∣2−2Re(zwˉ)|z - w|^2 = |z|^2 + |w|^2 - 2Re(z\bar{w})

Reveal Answer

RTP
∣z−w∣2=∣z∣2+∣w∣2−2Re(zw‾)|z - w|^2 = |z|^2 + |w|^2 - 2\text{Re}(z \overline{w})

LHS =∣z−w∣2= |z - w|^2
=∣(a+bi)−(c+di)∣2= |(a + bi) - (c + di)|^2
=∣(a−c)+(b−d)i∣2= |(a - c) + (b - d)i|^2
=(a−c)2+(b−d)2= (a - c)^2 + (b - d)^2
=a2−2ac+c2+b2−2bd+d2= a^2 - 2ac + c^2 + b^2 - 2bd + d^2

RHS =∣z∣2+∣w∣2−2Re(zw‾)= |z|^2 + |w|^2 - 2\text{Re}(z \overline{w})
=∣a+bi∣2+∣c+di∣2…= |a + bi|^2 + |c + di|^2 \dots
⋯−2Re((a+bi)(c−di))\dots - 2\text{Re}((a + bi)(c - di))
=a2+b2+c2+d2…= a^2 + b^2 + c^2 + d^2 \dots
⋯−2Re((ac+bd)+(bc−ad)i)\dots - 2\text{Re}((ac + bd) + (bc - ad)i)
=a2+b2+c2+d2−2(ac+bd)= a^2 + b^2 + c^2 + d^2 - 2(ac + bd)
=a2+b2+c2+d2−2ac−2bd= a^2 + b^2 + c^2 + d^2 - 2ac - 2bd
=a2−2ac+c2+b2−2bd+d2= a^2 - 2ac + c^2 + b^2 - 2bd + d^2
=LHS= \text{LHS}

Marking Criteria
DescriptorMarks

correctly expresses z−wz - w in terms of a,b,ca, b, c and dd in Cartesian form

1

expresses ∣z−w∣2|z - w|^2 in terms of a,b,ca, b, c and dd in expanded form

1

correctly expresses ∣z∣2+∣w∣2|z|^2 + |w|^2 in terms of a,b,ca, b, c and dd in expanded form

1

correctly expresses (zw‾)(z \overline{w}) in terms of a,b,ca, b, c and dd

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⁡(θ)+isin⁡(θ)))n=rn(cos⁡(nθ)+isin⁡(nθ))∀n∈Z+(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=1n = 1.
LHS=r(cos⁡(θ)+isin⁡(θ))\text{LHS} = r(\cos(\theta) + i\sin(\theta))
RHS=r1(cos⁡(1×θ)+isin⁡(1×θ))=r(cos⁡(θ)+isin⁡(θ))=LHS\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=kn = k.

(r(cos⁡(θ)+isin⁡(θ)))k=rk(cos⁡(kθ)+isin⁡(kθ))(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+1n = k + 1.

(r(cos⁡(θ)+isin⁡(θ)))k+1=rk+1(cos⁡((k+1)θ)+isin⁡((k+1)θ))(r(\cos(\theta) + i\sin(\theta)))^{k+1} = r^{k+1}(\cos((k+1)\theta) + i\sin((k+1)\theta)) LHS=(r(cos⁡(θ)+isin⁡(θ)))k⋅r(cos⁡(θ)+isin⁡(θ))1=rk(cos⁡(kθ)+isin⁡(kθ)) r(cos⁡(θ)+isin⁡(θ))=rk+1(cos⁡(kθ)cos⁡(θ)+icos⁡(kθ)sin⁡(θ)+isin⁡(kθ)cos⁡(θ)−sin⁡(kθ)sin⁡(θ))=rk+1(cos⁡(kθ)cos⁡(θ)−sin⁡(kθ)sin⁡(θ)+i(sin⁡(kθ)cos⁡(θ)+cos⁡(kθ)sin⁡(θ)))=rk+1(cos⁡(kθ+θ)+i(sin⁡(kθ+θ)))=rk+1(cos⁡((k+1)θ)+isin⁡((k+1)θ))=RHS\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+1n = k + 1. By mathematical induction, the rule is true for n=1,2,…n = 1, 2, \dots

Marking Criteria
DescriptorMarks

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.

∑r=1nr(r+1)=13n(n+1)(n+2)∀n∈Z+\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

∑r=1kk(k+1)=13k(k+1)(k+2)\sum_{r=1}^k k(k+1) = \frac{1}{3}k(k+1)(k+2)

B

∑r=1kk(k+1)=13n(n+1)(n+2)\sum_{r=1}^k k(k+1) = \frac{1}{3}n(n+1)(n+2)

C

∑r=1kr(r+1)=13k(k+1)(k+2)\sum_{r=1}^k r(r+1) = \frac{1}{3}k(k+1)(k+2)

D

∑r=1kr(r+1)=13n(n+1)(n+2)\sum_{r=1}^k r(r+1) = \frac{1}{3}n(n+1)(n+2)

Reveal Answer
A

∑r=1kk(k+1)=13k(k+1)(k+2)\sum_{r=1}^k k(k+1) = \frac{1}{3}k(k+1)(k+2)

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

B

∑r=1kk(k+1)=13n(n+1)(n+2)\sum_{r=1}^k k(k+1) = \frac{1}{3}n(n+1)(n+2)

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

C

∑r=1kr(r+1)=13k(k+1)(k+2)\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=kn=k by replacing every instance of nn in the original proposition with kk.

D

∑r=1kr(r+1)=13n(n+1)(n+2)\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 kk, the right-hand side still uses nn. The inductive hypothesis requires substituting kk for nn on both sides of the equation.

Q15
2023
QCAA
Paper 1
5 marks
Q15

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

1+r+r2+r3+...+rn−1=rn−1r−1(for r≠1).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 ∀n∈Z+\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=1n=1.
LHS=1LHS = 1
RHS=r−1r−1RHS = \frac{r-1}{r-1}
=1=LHS= 1 = LHS

Marking Criteria
DescriptorMarks

Correctly proves the initial statement

1
Q15b
3 marks

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

1+r+r2+r3+...+rk−1=rk−1r−1(r≠1).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+r2+r3+...+rk−1=rk−1r−1(r≠1)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+1n=k+1 for r≠1r \neq 1
1+r+r2+r3+...+rk−1+rk=rk+1−1r−11 + r + r^2 + r^3 + ... + r^{k-1} + r^k = \frac{r^{k+1}-1}{r-1}

LHS=rk−1r−1+rkLHS = \frac{r^k-1}{r-1} + r^k
=rk−1+rk+1−rkr−1= \frac{r^k-1 + r^{k+1} - r^k}{r-1}
=rk+1−1r−1= \frac{r^{k+1}-1}{r-1}
=RHS= RHS

Marking Criteria
DescriptorMarks

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

Marking Criteria
DescriptorMarks

States a suitable conclusion to the proof

1
Q2
2024
VCAA
Paper 1
3 marks
Q2
3 marks

Prove that if xx is an odd integer then 2x2−3x−72x^2 - 3x - 7 is even, using a direct proof.

Reveal Answer

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

2x2−3x−7=2(2k+1)2−3(2k+1)−7=2(4k2+4k+1)−6k−3−7=8k2+2k−8=2(4k2+k−4) which is even\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
DescriptorMarks

Provides a complete and correct proof using algebraic substitution, demonstrating that the expression factors to a multiple of 2

3

Correctly substitutes x=2k+1x = 2k + 1 and expands the expression, but provides an incomplete proof or makes a minor algebraic error

2

Substitutes x=2k+1x = 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+cis(θ))(1 + \text{cis}(\theta)) for n∈Z+n \in Z^+, (1+cis(θ))≠0(1 + \text{cis}(\theta)) \neq 0.

∑r=02n+1cis(rθ)\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
∑r=02n+1cis(rθ)\sum_{r=0}^{2n+1} \text{cis}(r\theta) is divisible by (1+cis(θ))(1 + \text{cis}(\theta)) for n∈Z+n \in \mathbb{Z}^+
Let n=1n = 1
∑r=03cis(rθ)=cis(0)+cis(θ)+cis(2θ)+cis(3θ)\sum_{r=0}^{3} \text{cis}(r\theta) = \text{cis}(0) + \text{cis}(\theta) + \text{cis}(2\theta) + \text{cis}(3\theta)
=1+cis(θ)+(cis(θ))2+(cis(θ))3= 1 + \text{cis}(\theta) + (\text{cis}(\theta))^2 + (\text{cis}(\theta))^3
=(1+cis(θ))(1+(cis(θ))2)= (1 + \text{cis}(\theta))(1 + (\text{cis}(\theta))^2)
This expression is divisible by (1+cis(θ))(1 + \text{cis}(\theta)), so the proposition is true for n=1n = 1.

Assume n=kn = k is true for k∈Z+k \in \mathbb{Z}^+:
∑r=02k+1cis(rθ)=(1+cis(θ))Q(θ)\sum_{r=0}^{2k+1} \text{cis}(r\theta) = (1 + \text{cis}(\theta))Q(\theta) where Q(θ)Q(\theta) is a function of θ\theta

Let n=k+1n = k + 1
∑r=02k+3cis(rθ)=∑r=02k+1cis(rθ)+cis((2k+2)θ)+cis((2k+3)θ)\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+cis(θ))Q(θ)+(cis(θ))2k+2+(cis(θ))2k+3= (1 + \text{cis}(\theta))Q(\theta) + (\text{cis}(\theta))^{2k+2} + (\text{cis}(\theta))^{2k+3}
=(1+cis(θ))Q(θ)+(cis(θ))2k+2(1+cis(θ))= (1 + \text{cis}(\theta))Q(\theta) + (\text{cis}(\theta))^{2k+2}(1 + \text{cis}(\theta))
=(1+cis(θ))(Q(θ)+(cis(θ))2k+2)= (1 + \text{cis}(\theta))(Q(\theta) + (\text{cis}(\theta))^{2k+2})
=(1+cis(θ))R(θ)= (1 + \text{cis}(\theta))R(\theta) where R(θ)R(\theta) is a function of θ\theta
The proposition is true for n=k+1n = k + 1. By mathematical induction, the formula is true for n=1,2,...n = 1, 2, ...

Marking Criteria
DescriptorMarks

correctly proves the initial statement

1

correctly establishes an appropriate assumption for n=kn=k

1

expresses the sum based on n=k+1n=k+1 in terms of the assumption

1

expresses a result using a common factor of (1+cis(θ))(1 + \text{cis}(\theta))

1

proves the inductive step

1
Q8
2021
QCAA
Paper 1
1 mark
Q8
1 mark

Let P(n)P(n) be the proposition that

∑r=1n(r+1)3r−1=n×3n∀n∈Z+\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)P(k) is true ∀k∈Z+\forall k \in Z^+ in a proof using mathematical induction?

A

∑r=1k(k+1)3k−1=k×3k\sum_{r=1}^k (k+1)3^{k-1} = k \times 3^k

B

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

C

∑r=1k(r+1)3r−1=k×3k\sum_{r=1}^k (r+1)3^{r-1} = k \times 3^k

D

∑r=1k(r+1)3r−1=r×3r\sum_{r=1}^k (r+1)3^{r-1} = r \times 3^r

Reveal Answer
A

∑r=1k(k+1)3k−1=k×3k\sum_{r=1}^k (k+1)3^{k-1} = k \times 3^k

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

B

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

This option incorrectly replaces rr with kk inside the sum and fails to replace nn with kk on the right-hand side of the equation.

C

∑r=1k(r+1)3r−1=k×3k\sum_{r=1}^k (r+1)3^{r-1} = k \times 3^k

Correct Answer

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

D

∑r=1k(r+1)3r−1=r×3r\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 kk, not the dummy index variable rr.

Q8
2023
VCAA
Paper 1
4 marks
Q8
4 marks

A function ff has the rule f(x)=xe2xf(x) = x e^{2x}.

Use mathematical induction to prove that f(n)(x)=(2nx+n2n−1)e2xf^{(n)}(x) = (2^n x + n 2^{n-1}) e^{2x} for n∈Z+n \in Z^+, where f(n)(x)f^{(n)}(x) represents the nthn^{\text{th}} derivative of f(x)f(x). That is, f(x)f(x) has been differentiated nn times.

Reveal Answer

f′(x)=e2x+2xe2x=(2x+1)e2x=(21x+1×21−1)e2x\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=1n=1.

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

f(k)(x)=(2kx+k2k−1)e2xf^{(k)}(x) = (2^kx + k2^{k-1})e^{2x}

then

f(k+1)(x)=ddx((2kx+k2k−1)e2x)=2ke2x+2(2kx+k2k−1)e2x=(2k+1x+(k+1)2k)e2x\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+1n = k+1.
Therefore, by the principle of mathematical induction, the statement is true for all n∈Z+n \in Z^+.

Marking Criteria
DescriptorMarks

Shows the statement is true for n=1n=1 by correctly finding f′(x)f'(x) and showing it matches the formula

1

States the inductive assumption that the statement is true for n=kn=k

1

Correctly differentiates f(k)(x)f^{(k)}(x) with respect to xx to find f(k+1)(x)f^{(k+1)}(x)

1

Simplifies the expression to show it matches the formula for n=k+1n=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 ∑j=1n(2j−1)=n2 ∀n∈Z+\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+1n=k+1 could be expressed as

A

∑j=1k+1(2j−1)=k2+2k+1\sum_{j=1}^{k+1} (2j-1) = k^2 + 2k + 1

B

∑j=1k+1(2k+1)=k2+2k+1\sum_{j=1}^{k+1} (2k+1) = k^2 + 2k + 1

C

∑j=1k+1(2j−1)=k2+1\sum_{j=1}^{k+1} (2j-1) = k^2 + 1

D

∑j=1k+1(2k+1)=k2+1\sum_{j=1}^{k+1} (2k+1) = k^2 + 1

Reveal Answer
A

∑j=1k+1(2j−1)=k2+2k+1\sum_{j=1}^{k+1} (2j-1) = k^2 + 2k + 1

Correct Answer

To express the proposition for n=k+1n=k+1, substitute k+1k+1 for nn on both sides of the equation. The upper limit of the sum becomes k+1k+1, and the right side becomes (k+1)2(k+1)^2, which expands to k2+2k+1k^2 + 2k + 1.

B

∑j=1k+1(2k+1)=k2+2k+1\sum_{j=1}^{k+1} (2k+1) = k^2 + 2k + 1

The term inside the summation must remain a function of the index variable jj (specifically 2j−12j-1). Replacing it with 2k+12k+1 incorrectly makes the term constant with respect to the summation index.

C

∑j=1k+1(2j−1)=k2+1\sum_{j=1}^{k+1} (2j-1) = k^2 + 1

The right side of the equation represents n2n^2 evaluated at n=k+1n=k+1. This should be (k+1)2=k2+2k+1(k+1)^2 = k^2 + 2k + 1, not k2+1k^2 + 1.

D

∑j=1k+1(2k+1)=k2+1\sum_{j=1}^{k+1} (2k+1) = k^2 + 1

This option is incorrect on both sides: the summation term should remain (2j−1)(2j-1), and the right side should be the expansion of (k+1)2(k+1)^2, which is k2+2k+1k^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)n(2n-1)(2n+1) is divisible by 3∀n∈Z+3 \forall n \in Z^+, the inductive step requires proving

A

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

B

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

C

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

D

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

Reveal Answer
A

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

This option incorrectly calculates the terms involving nn. Substituting n=k+1n=k+1 into (2n−1)(2n-1) yields 2k+12k+1, not 2k2k, and into (2n+1)(2n+1) yields 2k+32k+3, not 2k+22k+2.

B

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

The middle term is incorrect. When substituting n=k+1n=k+1 into (2n−1)(2n-1), the result is 2(k+1)−1=2k+12(k+1)-1 = 2k+1, not 2k2k.

C

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

The final term is incorrect. When substituting n=k+1n=k+1 into (2n+1)(2n+1), the result is 2(k+1)+1=2k+32(k+1)+1 = 2k+3, not 2k+22k+2.

D

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

Correct Answer

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

Q1
2024
VCAA
Paper 2
1 mark
Q1
1 mark

Consider the statement
'for any integers mm and nn, if m+n≥9m + n \ge 9 then m≥5m \ge 5 or n≥5n \ge 5'.
The contrapositive of this statement is

A

if m<5m < 5 or n<5n < 5, then m+n<9m + n < 9

B

if m≥5m \ge 5 or n≥5n \ge 5, then m+n≥9m + n \ge 9

C

if m<5m < 5 and n<5n < 5, then m+n<9m + n < 9

D

if m≤5m \le 5 and n≤5n \le 5, then m+n≤9m + n \le 9

Reveal Answer
A

if m<5m < 5 or n<5n < 5, then m+n<9m + n < 9

This option incorrectly applies De Morgan's laws by using 'or' instead of 'and' when negating the original conclusion.

B

if m≥5m \ge 5 or n≥5n \ge 5, then m+n≥9m + n \ge 9

This is the converse of the original statement, formed by swapping the hypothesis and conclusion without negating them.

C

if m<5m < 5 and n<5n < 5, then m+n<9m + n < 9

Correct Answer

The contrapositive negates and swaps the hypothesis and conclusion. By De Morgan's laws, the negation of 'm≥5m \ge 5 or n≥5n \ge 5' is 'm<5m < 5 and n<5n < 5', and the negation of 'm+n≥9m + n \ge 9' is 'm+n<9m + n < 9'.

D

if m≤5m \le 5 and n≤5n \le 5, then m+n≤9m + n \le 9

The negation of a 'greater than or equal to' (≥\ge) inequality is strictly 'less than' (<<), not 'less than or equal to' (≤\le).

Q4
2022
QCAA
Paper 1
1 mark
Q4
1 mark

When using proof by mathematical induction to prove De Moivre's theorem expressed as (rcis(θ))n=rncis(nθ)∀n∈Z+(\text{rcis}(\theta))^n = r^n \text{cis}(n\theta) \forall n \in Z^+, which statement would be correct in the proof of the inductive step?

A

(rcis(θ))k=rkcis(kθ)(\text{rcis}(\theta))^k = r^k \text{cis}(k\theta)

B

(rcis(θ))k=rk+1cis(k+θ)(\text{rcis}(\theta))^k = r^{k+1} \text{cis}(k+\theta)

C

(rcis(θ))k+1=rk+1cis(kθ+1)(\text{rcis}(\theta))^{k+1} = r^{k+1} \text{cis}(k\theta+1)

D

(rcis(θ))k+1=rk+1cis((k+1)θ)(\text{rcis}(\theta))^{k+1} = r^{k+1} \text{cis}((k+1)\theta)

Reveal Answer
A

(rcis(θ))k=rkcis(kθ)(\text{rcis}(\theta))^k = r^k \text{cis}(k\theta)

This represents the inductive hypothesis (assuming the theorem is true for n=kn=k). While used in the proof, the goal of the inductive step is to establish the validity of the formula for n=k+1n=k+1.

B

(rcis(θ))k=rk+1cis(k+θ)(\text{rcis}(\theta))^k = r^{k+1} \text{cis}(k+\theta)

This equation is incorrect because the exponent of rr on the right side does not match the power on the left, and the argument of the cosine/sine functions is algebraically wrong.

C

(rcis(θ))k+1=rk+1cis(kθ+1)(\text{rcis}(\theta))^{k+1} = r^{k+1} \text{cis}(k\theta+1)

This is incorrect because the argument inside the cis\text{cis} function follows the pattern nθn\theta; for n=k+1n=k+1, the argument must be (k+1)θ(k+1)\theta, not kθ+1k\theta + 1.

D

(rcis(θ))k+1=rk+1cis((k+1)θ)(\text{rcis}(\theta))^{k+1} = r^{k+1} \text{cis}((k+1)\theta)

Correct Answer

This correctly represents the theorem for n=k+1n=k+1. The inductive step requires proving that this equality holds, given the assumption that the theorem holds for n=kn=k.

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