Assignment 27

course Mth 151

MgҳŒ]assignment #027

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

wEZl}ɆרyO®Ï

Liberal Arts Mathematics I

07-24-2006

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11:20:52

5.5.6 Give the approximate value of Golden Ratio to thousandth and show how you obtained your result.

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RESPONSE -->

1.618

I just used my calculator and figured the equation (1 + square root of 5)/2

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11:21:02

** 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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RESPONSE -->

ok

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11:23:13

5.5.12 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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RESPONSE -->

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

21^3 + 13^3 - 8^3 = 10,946

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11:23:53

** 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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RESPONSE -->

ok

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11:26:23

5.5.18 show whether F(p+1) or F(p-1) is divisible by p.

Give your solution to this problem.

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RESPONSE -->

I don't think I got this correct, but here is what I tried.

p = 3

F(p+1), F(3+1), F(4)

4/3

F(p-1), F(3-1), F(2)

2/4

That's about as far as I got.

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11:29:38

** 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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RESPONSE -->

In the very first sentence, how does f(2) = 1? Does f stand for the number in the Fib. sequence? The second number in the sequence is 1?

Okay, I got it. I read the rest of the examples.

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11:31:00

5.5.24 Lucas sequence: L2 + L4; L2 + L4 + L6; etc.. Give your solution to this problem as stated in your text.

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RESPONSE -->

3 + 7 = 10

3 + 7 + 18 = 28

3 + 7 + 18 + 47 = 75

3 + 7 + 18 + 47 + 123 = 198

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11:32:57

** 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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RESPONSE -->

I did the equation, but I didn't see the pattern. I see it now.

.................................................

"

I think you understand this. Let me know if you have questions.

Assignment 27

course Mth 151

MgҳŒ]assignment #027

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

wEZl}ɆרyO®Ï

Liberal Arts Mathematics I

07-24-2006

......!!!!!!!!...................................

11:20:52

5.5.6 Give the approximate value of Golden Ratio to thousandth and show how you obtained your result.

......!!!!!!!!...................................

RESPONSE -->

1.618

I just used my calculator and figured the equation (1 + square root of 5)/2

.................................................

......!!!!!!!!...................................

11:21:02

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

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

......!!!!!!!!...................................

RESPONSE -->

ok

.................................................

......!!!!!!!!...................................

11:23:13

5.5.12 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?

......!!!!!!!!...................................

RESPONSE -->

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

21^3 + 13^3 - 8^3 = 10,946

.................................................

......!!!!!!!!...................................

11:23:53

** 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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RESPONSE -->

ok

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11:26:23

5.5.18 show whether F(p+1) or F(p-1) is divisible by p.

Give your solution to this problem.

......!!!!!!!!...................................

RESPONSE -->

I don't think I got this correct, but here is what I tried.

p = 3

F(p+1), F(3+1), F(4)

4/3

F(p-1), F(3-1), F(2)

2/4

That's about as far as I got.

.................................................

......!!!!!!!!...................................

11:29:38

** 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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RESPONSE -->

In the very first sentence, how does f(2) = 1? Does f stand for the number in the Fib. sequence? The second number in the sequence is 1?

Okay, I got it. I read the rest of the examples.

.................................................

......!!!!!!!!...................................

11:31:00

5.5.24 Lucas sequence: L2 + L4; L2 + L4 + L6; etc.. Give your solution to this problem as stated in your text.

......!!!!!!!!...................................

RESPONSE -->

3 + 7 = 10

3 + 7 + 18 = 28

3 + 7 + 18 + 47 = 75

3 + 7 + 18 + 47 + 123 = 198

.................................................

......!!!!!!!!...................................

11:32:57

** 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. **

......!!!!!!!!...................................

RESPONSE -->

I did the equation, but I didn't see the pattern. I see it now.

.................................................

"

I think you understand this. Let me know if you have questions.