#$&* course mth151 12/17/14 2 If your solution to stated problem does not match the given solution, you should self-critique per instructions
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Given Solution: `a** The Golden Ratio is [ 1+`sqrt(5) ] /2 [ 1+`sqrt(5) ] /2=[ 1+2.2361 ] /2=3.2361/2=1.618 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q5.4.12 2^3 + 1^3 - 1^3 = 8; 3^3 + 2^3 - 1^3 = 34; 5^3 + 3^3 - 2^3 = 144; 8^3 + 5^3 - 3^3 = 610. What are the next two equations in this sequence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2^3+1^3-1^3=8 3^3+2^3-1^3=34 5^3+3^3-2^3=144 8^3+5^3-3^3=610 13^3+8^3-5^3=2584 21^3+13^3-8^3=10946 fibonacci sequence 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610 notice the results starting at 8 and then skip 2 to the next result next sequence should be equal to three numbers down the line;2584 each part of the equation is a number along the fibonacci sequence and with each equation, it goes up on the fibonacci sequence by 1 ""term"";the first part of the equation starts at 2 then moves to 3, then 5,8,then 13,21,ect... 377+610=987 610+987=1597 987+1597=2584 2584+1597=4181 2584+4181=6765 4181+6765=10946 confidence rating #$&*: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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q5.4.18 show whether F(p+1) or F(p-1) is divisible by p. Give your solution to this problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** For p=3 we get f(p-1) = f(2) = 1 and f(p+1) = f(4)= 3; f(p+1) = f(4) = 3 is divisible by p, which is 3 So the statement is true for p = 3. For p=7 we get f(p-1) = f(6) = 8 and f(p+1) = f(8) = 21; f(p+1) = 21 is divisible by p = 7. So the statement is true for p = 7. For p = 11 we get f(p-1) = f(10)= 55 and f(p+1) = f(12) = 144. f(p-1) = 55 is divisible by p = 11. So the statement is true for p = 11. So the conjecture is true for p=3, p=7 and p=11.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q5.4.24 Lucas sequence: L2 + L4; L2 + L4 + L6; etc.. Give your solution to this problem as stated in your text. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: luca sequence 1.3.4.7.11.18.29.47.76.123 L2+L4 = 3+7=10 L2+L4+L6 = 3+7+18=28 L2+L4+L6+L8 = 3+7+18+47=75 L2+L4+L6+L8+L10 = 3+7+18+47+123=198 10 is 1 less than L5 28 is 1 less than L7 75 is 1 less than L9 198 is 1 less than L11 which means... L2+L4=L5-1 L2+L4+L6=L7-1 L2+L4+L6+L8=L9-1 L2+L4+L6+L8+L10=L11-1 confidence rating #$&*: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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!