How many QCAA Specialist Mathematics questions cover Mathematical induction and trigonometric proofs?

Q: How many QCAA Specialist Mathematics questions cover Mathematical induction and trigonometric proofs?

AusGrader has 19 QCAA Specialist Mathematics questions on Mathematical induction and trigonometric proofs, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Specialist Mathematics — Mathematical induction and trigonometric proofs Questions & Answers | AusGrader

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

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

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

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

Q12

2023

QCAA

Paper 2

7 marks

Q12

Consider the complex number $z = -3 + 2i$ .

Q12a

2 marks

Determine $z^3$ using the binomial theorem. Leave your answer in the form $a + bi$ , where $a, b \in R$ .

Descriptor	Marks
correctly uses the binomial theorem	1
expresses $z^3$ in the form of $a+bi$	1

Q12b

1 mark

Convert $z$ into the form $r \text{ cis}(\theta)$ , where $-\pi < \theta \leq \pi$ .

$z = 3.61\text{cis}(2.55)$

Marking Criteria

Descriptor	Marks
correctly converts $z$ into $r\text{cis}(\theta)$ form where $-\pi < \theta \le \pi$	1

Q12c

2 marks

Use the result from Question 12b) to determine $z^3$ using De Moivre's theorem. Leave your answer in the form $r \text{ cis}(\theta)$ , where $-\pi < \theta \leq \pi$ .

Using De Moivre's theorem
$z^3 = (3.61\text{cis}(2.55))^3$
$= 3.61^3\text{cis}(3 \times 2.55)$
$= 46.87\text{cis}(7.66)$
$= 46.87\text{cis}(1.38)$

Marking Criteria

Descriptor	Marks
uses De Moivre's theorem to calculate $z^3$	1
determines $z^3$ in the form of $r\text{cis}(\theta)$ where $-\pi < \theta \le \pi$	1

Q12d

2 marks

Evaluate the reasonableness of your results from Questions 12a) and 12c), noting that the two methods to determine $z^3$ should produce the same result.

$46.87\text{cis}(1.38) = 46.87(\cos(1.38) + i\sin(1.38))$
$= 8.89 + 46.02i$
$\approx 9 + 46i$

The two methods produce approximately the same $z^3$ value. The small variation is a result of rounding used in earlier calculations.

Marking Criteria

Descriptor	Marks
shows mathematical reasoning to convert from $r\text{cis}(\theta)$ form to $a+bi$ form	1
states a decision regarding the reasonableness	1

Q16

2021

QCAA

Paper 1

6 marks

Q16

6 marks

Use mathematical induction to prove that $2^{2n} + 3n - 1$ is divisible by $3 \forall n \in Z^+$ .

RTP
$2^{2n} + 3n - 1$ is always divisible by 3 for $n \in \mathbb{Z}^+$ ,
i.e. $2^{2n} + 3n - 1 = 3m$ , for some $m \in \mathbb{Z}^+$
Initial statement: Let $n = 1$ :
$2^2 + 3 - 1 = 6$
$= 3 \times 2$
Proposition is true for $n = 1$

Assume proposition is true for $n = k \forall k \in \mathbb{Z}^+$
$2^{2k} + 3k - 1 = 3m$ for some $m \in \mathbb{Z}^+$

Inductive step: Let $n = k + 1$
$2^{2(k+1)} + 3(k + 1) - 1 = 2^{2k+2} + 3k + 3 - 1$
$= 2^2 \times 2^{2k} + 3k - 1 + 3$
$= 2^{2k} + 3k - 1 + 3 \times 2^{2k} + 3$
$= 3m + 3 \times 2^{2k} + 3$
$= 3(m + 2^{2k} + 1)$
$= 3p$ for some $p \in \mathbb{Z}^+$

The proposition is true for $n = k + 1$
By mathematical induction, the proposition is true for $n = 1, 2, \dots$

Marking Criteria

Descriptor	Marks
correctly proves the initial statement	1
correctly formulates an assumption for $n = k$	1
correctly establishes an expression representing the LHS of the inductive step proof and uses an index law to establish a term with a factor of $2^{2k}$	1
uses previous assumption in the inductive step	1
completes proof by determining an expression with a factor of 3 representing the RHS of the inductive step proof	1
shows logical organisation, having attempted all steps of the proof, including the use of a suitable conclusion	1

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:

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:

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:

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

QCAA Specialist Mathematics Mathematical induction and trigonometric proofs

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