course Mth 151 Could not locate q_a_ 27. SyXoڐٚxR
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21:46:08 5.5.6 Give the approximate value of Golden Ratio to thousandth and show how you obtained your result.
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RESPONSE --> 1.618033989 rounded to the thousandth is 1.618
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21:47:00 ** 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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21:54:45 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 = 10946
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22:04:44 ** 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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22:19:36 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 --> a) 3 F 2 + 1 F 2 -1 3 1 F p + 1 is / by p b) 7 F 13 + 1 F 13 - 1 14 12 F p + 1 is / by p c) 11 F 89 + 1 F 89 - 1 90 88 F p - 1 is / by p
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22:24:48 ** 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 --> This is how I did it the first time but wasn't sure if it was right so I substituted the Fibonacci numbers for the value of p and solved.
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22:40:13 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 --> 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 The differences between the equations are the last number in the sequence. 198 - 75 = 123 or L10, 75 - 28 = 47 or L8, 28 - 10 = 18 or L6.
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22:47:16 ** 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 see the sequence now.
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