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anonymous

  • one year ago

a12 and a20 are the 12th and 20th terms of Fibonacci sequence.find (a12,a20)

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  1. anonymous
    • one year ago
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    first i wrote the first terms of Fibonacci sequence. 1,1,2,3,5,8,13,21,34,55,89,144,233 then i solved these examples : (a2,a4)=(1,3)=1=a2 (a3,a6)=(2,8)=2=a3 (a4,a6)=(3,8)=1=a2 so i guessed (am,an)=a(m,n)

  2. amistre64
    • one year ago
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    what does: find (a12,a20) mean?

  3. anonymous
    • one year ago
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    (a,b) means Greatest common divisor of a and b

  4. amistre64
    • one year ago
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    hmm, you only wroted 13 terms .... any reason you didnt just compute it with the actual values?

  5. amistre64
    • one year ago
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    1 = a1 or a2

  6. amistre64
    • one year ago
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    not sure if there is a way i can think of to prove your conjecture tho ... unless we develop the closed form of a fibber

  7. amistre64
    • one year ago
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    x^(n+2) = x^(n+1) + x^n x^(n+2) - x^(n+1) - x^n = 0 x^n (x^2 - x - 1) = 0 x = 1/2 (1 +- sqrt(5)) ................................. setting a system we get Ax1^0 + Bx2^0 = 1 , since f(0) = 1 Ax1^1 + Bx2^1 = 1 , since f(1) = 1 ------------------ A + B = 1 Ax1 + Bx2 = 1 -Ax1 - Bx1 = -x1 Ax1 + Bx2 = 1 ---------------- B(x2-x1) = 1-x1 B = (1-x1)/(x2-x1) B = (1-(1 + sqrt(5))/2)/((1 - sqrt(5))/2-(1 + sqrt(5))/2) B = (5-sqrt(5))/10 and A = 1- B A = (5+sqrt(5))/10 the closed form is therefore: Ax1^n + Bx2^n

  8. amistre64
    • one year ago
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    of course im starting at n=0 .. so we would have to adjust for index

  9. amistre64
    • one year ago
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    not sure if this would be useful in a GCD setup tho ... since its 2terms

  10. amistre64
    • one year ago
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    well, it would at least be useful in defining the nth terms for large values of n, but then so could coding a computer script to work the manual stuff for you.

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