To prove: f n is a natural number for all natural numbers n .

Question

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Answer 1

Question

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

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2. Consider the ODE u' = ƒ (u) = u² + r where r is a parameter that can take the values r = −1, −0.5, -0.1, 0.1. For each value of r: (a) Sketch ƒ(u) = u² + r and determine the equilibrium points. (b) Draw the phase line. (d) Determine the stability of the equilibrium points. (d) Plot the direction field and some sample solutions,i.e., u(t) (e) Describe how location of the equilibrium points and their stability change as you increase the parameter r. (f) Using the matlab program phaseline.m generate a solution for each value of r and the initial condition u(0) = 0.9. Print and turn in your result for r = −1. Do not forget to add a figure caption. (g) In the matlab program phaseline.m set the initial condition to u(0) = 1.1 and simulate the ode over the time interval t = [0, 10] for different values of r. What happens? Why? You do not need to turn in a plot for (g), just describe what happens.

True or False and why

Answer 2

Question

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

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Answer 3

Question

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

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Answer 4

Question

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

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Answer 5

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

Answer 6

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

Answer 7

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

Answer 8

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

Answer 13

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

Answer 14

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

Answer 19

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

Answer 20

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

Answer 25

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Answer 26

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .