54 qa 2

#$&*

course Mth 151

153 pm 12/2/12

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:

1+ the square root of 5 /2

=1.618

confidence rating #$&*:

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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:3

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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:

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

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

confidence rating #$&*:

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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:

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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:

confidence rating #$&*:

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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):

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Self-critique Rating:

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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:

10, 28, 75, 198

It is always one less than a number in the Lucas sequence.

confidence rating #$&*:

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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:

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

10, 28, 75, 198

It is always one less than a number in the Lucas sequence.

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:

#*&!

&#Very good responses. Let me know if you have questions. &#