course Mth 151
April 25 around 9:40
assignment 27-QUERYq5.5.2:
The golden ratio would be [1+'sqrt(5)]/2--this now looks like this in the equation:[1+`sqrt(5)]/2=[1+2.2361]/2=3.2361/2=1.618
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q5.5.12:
The next equation would be f(7^)3+f(6)^3-f(5)^3=f(17). Then we do substituting and get 13^3+8^3-5^3=f(17). Therefore the left-handed side would give us the result as 2584.
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q5.5.18:
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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q5.5.24:
The Lucas sequence is 1 3 4 7 11 18 29 47 76 123 199etc.
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.
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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