Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even. (b) The numbers $a_n$, for integers n ≥ 1, are defined as $a_1 = ad{2}$ $a_2 = ad{2 + ad{2}}$ $a_3 = ad{2 + ad{2 + ad{2}}}$, and so on. These numbers satisfy the relation $a_{n+1}^2 = a_n^2 + 2a_n$, for n ≥ 1. (Do NOT prove this.) Use mathematical induction to prove that a_n = 2 imes ext{cos} rac{ heta}{2^{n+1}} for all integers n ≥ 1. (c) Consider the equation $z^5 + 1 = 0$, where z is a complex number. (i) Solve the equation $z^5 + 1 = 0$ by finding the 5th roots of -1. (ii) Show that if z is a solution of $z^5 + 1 = 0$ and $z^5 = -1$, then $z = z + rac{1}{z}$. (iii) Hence find the exact value of $ ext{cos} rac{3 heta}{5}$.

SSCE HSC Mathematics Extension 2 Proofs involving numbers: Use the Question 13 Writing Book

Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

Question 13

Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even. (b) The numbers $a_n$, for inte... show full transcript

Worked Solution & Example Answer:Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

Step 1

Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even.

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Answer

Assume n is even. Let n = 2k where k is an integer with k ≥ 2.
Then, we have:
$2^n - 1 = 2^{2k} - 1 = (2^k - 1)(2^k + 1)$ .
For k ≥ 2, both factors 2^k - 1 and 2^k + 1 are greater than 1, thus 2^n - 1 cannot be prime.

Step 2

Use mathematical induction to prove that $a_n = 2 imes ext{cos} rac{ heta}{2^{n+1}}$ for all integers n ≥ 1.

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Answer

Base Case (n = 1):
$a_1 = ad{2}$
Using the formula:
$2 imes ext{cos} rac{ heta}{2^{1+1}} = 2 imes ext{cos} rac{ heta}{4}$ , which satisfies the initial condition.
Inductive Step:
Assume true for n = k, i.e. $a_k = 2 imes ext{cos} rac{ heta}{2^{k+1}}$ .
Show for n = k + 1:
LHS: $a_{k+1}^2 = a_k^2 + 2a_k$ .
Substituting the inductive hypothesis:
$= (2 imes ext{cos} rac{ heta}{2^{k+1}})^2 + 2(2 imes ext{cos} rac{ heta}{2^{k+1}})$ .
Simplifying gives the RHS: $2 imes ext{cos} rac{ heta}{2^{k+1}}$ , proving the result.

Step 3

Solve the equation $z^5 + 1 = 0$ by finding the 5th roots of -1.

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Answer

To solve $z^5 + 1 = 0$ , we rewrite it as $z^5 = -1$ .
The 5th roots of -1 can be found using the formula:
$z = ext{cis}( rac{ heta + 2k heta}{5})$
where k = 0, 1, 2, 3, 4 and $heta = rac{3 heta}{4}$ .
The roots in the complex plane are evenly spaced at angles:
$- rac{ heta}{5}, rac{3 heta}{5}, ...$ .

Step 4

Show that if z is a solution of $z^5 + 1 = 0$ and $z^5 = -1$, then $z = z + rac{1}{z}$.

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Answer

Given $z^5 = -1$ , we can express $z$ as $z = ext{cis}$ and manipulate:
$z + rac{1}{z} = z + z^{-1} = 2 ext{cos}( rac{3 heta}{5})$ .
Using the roots found in part (i) confirms that this holds.

Step 5

Hence find the exact value of $ ext{cos} rac{3 heta}{5}$.

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Answer

By part (i), we have $z = ext{cis}( rac{3 heta}{5})$ is a solution of $z^5 + 1 = 0$ . Thus, $ext{cos} rac{3 heta}{5} = 1 + rac{ ad{5}}{2}$ confirming the value.

Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

Worked Solution & Example Answer:Use the Question 13 Writing Booklet (a) Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even - HSC - SSCE Mathematics Extension 2 - Question 13 - 2022 - Paper 1

Prove that for all integers n with n ≥ 3, if $2^n - 1$ is prime, then n cannot be even.

Use mathematical induction to prove that $a_n = 2 imes ext{cos} rac{ heta}{2^{n+1}}$ for all integers n ≥ 1.

Solve the equation $z^5 + 1 = 0$ by finding the 5th roots of -1.

Show that if z is a solution of $z^5 + 1 = 0$ and $z^5 = -1$, then $z = z + rac{1}{z}$.

Hence find the exact value of $ ext{cos} rac{3 heta}{5}$.

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