MMM #35: Is it too hard?

June 25th, 2009 | by Sol |

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.

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  1. 5 Responses to “MMM #35: Is it too hard?”

  2. By Brent Yorgey on Jun 25, 2009 | Reply

    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)!

  3. By efrique on Jun 26, 2009 | Reply

    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.

  4. By Alec on Jun 26, 2009 | Reply

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

  5. By papiauche on Jun 29, 2009 | Reply

    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)

    then

    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!

    then:

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

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