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 = √2 a_2 = √2 + √2 a_3 = √2 + √2 + √2, and so on. These numbers satisfy the relation a_{n+1}^2 = a_n^2 + 2a_n, for n ≥ 1. Use mathematical induction to prove that a_n = 2cos(π/(2n+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^k = -1, then z = z + 1/z. (iii) Hence find the exact value of cos(3π/5).

SSCE HSC Mathematics Extension 2 Roots of unity: 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 integers n ≥... 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.

96%

114 rated

Answer

To prove this statement, we will prove the contrapositive. Assume n is even, meaning n can be expressed as n = 2k for some integer k ≥ 2. Then:

2^n - 1 = 2^{2k} - 1 = (2^k - 1)(2^k + 1).

For k ≥ 2, both factors are greater than 1, hence 2^n − 1 has at least two proper factors and cannot be prime. Thus, we conclude that if 2^n - 1 is prime, then n must be odd.

Step 2

Use mathematical induction to prove that a_n = 2cos(π/(2n+1)) for all integers n ≥ 1.

99%

104 rated

Answer

We begin by checking the base case for n = 1:

a_1 = √2 = 2cos(π/3)

which holds true. Now, assume the relation holds for some n = k:

a_k = 2cos(π/(2k+1)).

We need to show it holds for n = k + 1:

a_{k+1}^2 = a_k^2 + 2a_k.

Substituting our inductive hypothesis gives:

display the relationship:

2cos(π/(2k+1))^2 + 2(2cos(π/(2k+1)))

After simplifying this expression using trigonometric identities, we find that:

a_{k+1} = 2cos(π/(2(k+1)+1))

Thus, by induction, the formula holds for all integers n ≥ 1.

Step 3

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

96%

101 rated

Answer

To find the roots of z^5 + 1 = 0, we can rewrite the equation as:

z^5 = -1

The 5th roots of -1 can be represented in exponential form:

z = e^{(2k + 1)πi/5}, ext{ for } k = 0, 1, 2, 3, 4.

This gives us the five roots, which are:

z = e^{πi/5}, e^{3πi/5}, e^{5πi/5}, e^{7πi/5}, e^{9πi/5}.

Step 4

Show that if z is a solution of z^5 + 1 = 0 and z^k = -1, then z = z + 1/z.

98%

120 rated

Answer

Assuming z is a solution of z^5 + 1 = 0 and z^k = -1, we assert:

u = z + rac{1}{z}$$ Calculating z + 1/z:

u = e^{(2k + 1)πi/5} + e^{-(2k + 1)πi/5} = 2cosigg(rac{(2k + 1)π}{5}igg)$$

We can conclude that:

u = z + rac{1}{z}$$ does hold.

Step 5

Hence find the exact value of cos(3π/5).

97%

117 rated

Answer

To find cos(3π/5), we can use the relationship derived earlier. From earlier calculations:

Using the angles we derived, we find:

cos(3π/5) = - rac{1}{2}

This leads us to the final value of cos(3π/5).

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 = 2cos(π/(2n+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^k = -1, then z = z + 1/z.

Hence find the exact value of cos(3π/5).

Join the SSCE students using SimpleStudy...