1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fibonacci Numbers

  1. Mar 23, 2009 #1
    1.Obtain the following formula for the alternating sum of Fibonacci Numbers

    f(1)-f(2)+f(3)-f(4)+...+(-1)^(n+1) f(n)= (-1)^(n+1) [f(n-1) ]+1

    2. Relevant equations

    3. I subtracted the formulas for the sum of the first n Fibonacci numbers with odd indices together with the first n Fibonacci numbers with even indices

    f(1)+f(3)+f(5)+...+f(2n-1) = f(2n)
    - [f(2)+f(4)+f(6)+...+f(2n) = [f(2n+1) ]- 1]

    = f(1)-f(2)+f(3)-f(4)+f(5)-f(6)+...+f(2n-1)-f(2n) = f(2n) - [f(2n+1)] +1

    what I am having trouble with is obtaining the (-1)^(n+1) factor
    I am thinking that if I factor the first two terms of the difference I would get the (-1) but as for its exponent I am not quite clear on how to derive it
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Discussions: Fibonacci Numbers
  1. Fibonacci Numbers (Replies: 7)

  2. Fibonacci Numbers (Replies: 1)

  3. Fibonacci numbers (Replies: 1)