assignment 27 query

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course mth151

12/17/14 2

If your solution to stated problem does not match the given solution, you should self-critique per instructions

at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase

interpretation of the problem along with a statement of what you do or do not understand about it. This response

should be given, based on the work you did in completing the assignment, before you look at the given solution.

027. `query 27

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Question: `q5.4.6 Give the approximate value of Golden Ratio to thousandth and show how you obtained your

result.

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Your solution:

golden ration is (1+ sqrt5)/2

(1+sqrt5)/2=(1+2.236)2=3.236/2=1.618

confidence rating #$&*:3

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Given Solution:

`a** The Golden Ratio is [ 1+`sqrt(5) ] /2

[ 1+`sqrt(5) ] /2=[ 1+2.2361 ] /2=3.2361/2=1.618 **

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q5.4.12 2^3 + 1^3 - 1^3 = 8; 3^3 + 2^3 - 1^3 = 34; 5^3 + 3^3 - 2^3 = 144; 8^3 + 5^3 - 3^3 = 610.

What are the next two equations in this sequence?

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Your solution:

2^3+1^3-1^3=8

3^3+2^3-1^3=34

5^3+3^3-2^3=144

8^3+5^3-3^3=610

13^3+8^3-5^3=2584

21^3+13^3-8^3=10946

fibonacci sequence

1,1,2,3,5,8,13,21,34,55,89,144,233,377,610

notice the results starting at 8 and then skip 2 to the next result

next sequence should be equal to three numbers down the line;2584

each part of the equation is a number along the fibonacci sequence and with each equation, it goes up on the

fibonacci sequence by 1 ""term"";the first part of the equation starts at 2 then moves to 3, then 5,8,then

13,21,ect...

377+610=987

610+987=1597

987+1597=2584

2584+1597=4181

2584+4181=6765

4181+6765=10946

confidence rating #$&*:3

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Given Solution:

`a** The numbers 1, 1, 2, 3, 5, 8, ... are the Fibonacci numbers f(1), f(2), ... The left-hand sides are

f(3)^3 + f(2)^3 - f(1)^3,

f(4)^3 + f(3)^3 - f(2)^3,

f(5)^3 + f(4)^3 - f(3)^3,

f(6)^3 + f(5)^3 - f(4)^3 etc..

The right-hand sides are f(5) = 8, f(8) = 34, f(11) = 144, f(14) = 610. So the equations are

f(3)^3 + f(2)^3 - f(1)^3 = f(5)

f(4)^3 + f(3)^3 - f(2)^3 = f(8)

f(5)^3 + f(4)^3 - f(3)^3 = f(11)

f(6)^3 + f(5)^3 - f(4)^3 = f(14)

etc..

The next equation would be

f(7)^3 + f(6)^3 - f(5)^3 = f(17).

Substituting f(7) = 13, f(6) = 8 and f(5) = 5 we get

13^3 + 8^3 - 5^3 = f(17). The left-hand side gives us result 2584, which is indeed f(17), so the pattern is

verified in this instance. **

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q5.4.18 show whether F(p+1) or F(p-1) is divisible by p.

Give your solution to this problem.

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Your solution:

p=3

f(p-1)=f(3-1)=f(2)=1

f(p+1)=f(3+1)=f(4)=3

3 is divisible by p so its true

p=7

f(p-1)=f(7-1)=f(6)=8

f(p+1)=f(7+1)=f(8)=21

21 is divisible by p so its true

p=11

f(p-1)=f(11-1)=f(10)=55

f(p+1)=f(11+1)=f(12)=144

55 is divisible by p so its true

confidence rating #$&*:3

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Given Solution:

`a** For p=3 we get f(p-1) = f(2) = 1 and f(p+1) = f(4)= 3; f(p+1) = f(4) = 3 is divisible by p, which is 3 So

the statement is true for p = 3.

For p=7 we get f(p-1) = f(6) = 8 and f(p+1) = f(8) = 21; f(p+1) = 21 is divisible by p = 7. So the statement is

true for p = 7.

For p = 11 we get f(p-1) = f(10)= 55 and f(p+1) = f(12) = 144. f(p-1) = 55 is divisible by p = 11. So the

statement is true for p = 11.

So the conjecture is true for p=3, p=7 and p=11.**

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q5.4.24 Lucas sequence: L2 + L4; L2 + L4 + L6; etc.. Give your solution to this problem as stated

in your text.

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Your solution:

luca sequence 1.3.4.7.11.18.29.47.76.123

L2+L4 = 3+7=10

L2+L4+L6 = 3+7+18=28

L2+L4+L6+L8 = 3+7+18+47=75

L2+L4+L6+L8+L10 = 3+7+18+47+123=198

10 is 1 less than L5

28 is 1 less than L7

75 is 1 less than L9

198 is 1 less than L11

which means...

L2+L4=L5-1

L2+L4+L6=L7-1

L2+L4+L6+L8=L9-1

L2+L4+L6+L8+L10=L11-1

confidence rating #$&*:3

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Given Solution:

`a** The Lucas sequence is 1 3 4 7 11 18 29 47 76 123 199 etc.

So

L2 + L4 = 3 + 7 = 10;

L2 + L4 + L6 = 3 + 7 + 18 = 28;

L2 + L4 + L6 + L8 = 3 + 7 + 18 + 47 = 75, and

L2 + L4 + L6 + L8 + L10 = 3 + 7 + 18 + 47 + 123 = 198.

Note that 10 is 1 less than 11, which is L5; 27 is 1 less than 28, which is L7; and 198 is 1 less than 199, which

is L9.

So L2 + L4 = L5 - 1, L2 + L4 + L6 = L7 - 1, etc..

So we can conjecture that the sum of a series of all evenly indexed members of the Lucas sequence, starting with

index 2 and ending with index 2n, is 1 less than member 2n+1 of the sequence. **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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