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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
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