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MMM #35: Is it too hard?

I've only gotten two solutions to the "Fibonacci Fun" problem I proposed. Is everyone on vacation and not thinking about Math, do people find the problem boring or is it too hard? I don't think it's one of my harder problems. I want to state that I approached the problem in a different way than each of the people who provided solutions. So, there are at least three ways to approach the problem.

If you don't know how to approach it I suggest looking for a pattern in the sum as you add more terms.

Come on, give it a try! You could win $10.

This time around, since I've given you guys a hint, I'll give three prizes - one each to the two people who have solved it already and one to a randomly other person with a correct solution.

Comments (4) Trackbacks (1)
  1. I just solved it, and I think it’s one of those tricky ones whose difficulty is hard to judge. It’s simple once you “see it”, but “seeing it” took me a while! Once I saw the pattern it was easy to prove by induction, but I’m not sure how you would work it out directly (without busting out heavy machinery like generating series)!

  2. I am sure I have a solution, but there’s no way I have time to write it up; very busy with other (work, mostly) things.

  3. I have a solution, I’m just lazy and on vacation. I’ll write it up and send it in before the deadline though.

  4. If n is even:
    S(n) = F(n)*F(n)

    If n is odd:
    S(n) = F(n)*F(n) – 1

    Demo: recurrence

    0 et 1 OK

    http://fr.wikipedia.org/wiki/Nombre_de_Fibonacci for Cassini identity

    Case n odd:

    S(n+1) = S(n) +F(n)*F(n+1) = F(n)*F(n) – 1 +F(n)*F(n+1)

    Cassini: F(n)*F(n) = F(n-1)*F(n+1) + 1

    S(n+1)= F(n-1)*F(n+1)+F(n)*F(n+1)= F(n+1)*(F(n-1)+F(n))

    Fibonacci relationship: F(n-1)+F(n) = F(n+1)


    S(n+1) = F(n+1)*F(n+1)

    Case n even

    S(n+1) = S(n) +F(n)*F(n+1) = F(n)*F(n)+F(n)*F(n+1)

    Cassini: F(n)*F(n) = F(n-1)*F(n+1) – 1

    S(n+1)= F(n-1)*F(n+1)+F(n)*F(n+1)-1= F(n+1)*(F(n-1)+F(n)) -1

    Fibonacci relationship:F(n-1)+F(n) = F(n+1)

    Comment: I’m French et poor english writer, sorry!


    S(n+1) = F(n+1)*F(n+1) -1

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