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Math Help - fibonacci series - a slight variant

  1. #1
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    fibonacci series - a slight variant

    Let me define a recurence like this

    S(1) = 1
    S(2) = 1

    S(n) = S(n-1) + S(n-2) + a; where a is some number

    Q1: What is the relation beween S(n) and F(n) [nth Fibonacci number]

    I found out this is S(n) = F(n) + K(n)*a

    Where K(n) is another series of the type S(n) where a = 1

    Q2: Is there a way to find the generic term for S(n) ?

    Any pointers will be of great help. Thanks
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  2. #2
    A Plied Mathematician
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    Well, you can always use the theory of difference equations (recurrence relations). You can find the nth term of the Fibonacci sequence this way, and I'm sure your S(n) sequence can be found the same way. The theory is very similar to that of differential equations.
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  3. #3
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    Let  f(n)  be a sequence that  f(n+2) = f(n+1) + f(n) , not necessarily be Fibonacci seq. , depends on the initial conditions .



    Then we have  S(n) = f(n) - a because

     f(n+2) - a = f(n+1) + f(n) - a = (f(n+1) - a) + (f(n) - a ) + a

     = S(n+1) + S(n) + a
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by aman_cc View Post
    Let me define a recurence like this

    S(1) = 1
    S(2) = 1

    S(n) = S(n-1) + S(n-2) + a; where a is some number

    Q1: What is the relation beween S(n) and F(n) [nth Fibonacci number]

    I found out this is S(n) = F(n) + K(n)*a

    Where K(n) is another series of the type S(n) where a = 1

    Q2: Is there a way to find the generic term for S(n) ?

    Any pointers will be of great help. Thanks
    We have  S_3=2+a, \; S_4=3+2a .

    You can use induction to show  S_n=F_n+a\cdot (F_n-1) .

    Edit: Credit goes to Soroban's post below. I had made a mistake.
    Last edited by chiph588@; July 8th 2010 at 09:07 AM.
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  5. #5
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    Hello, aman_cc!

    Let me define a recurence like this:

    . . S(1) = 1
    . . S(2) = 1
    . . S(n) \:= \:S(n\!-\!1) + S(n\!-\!2) + a\:\text{ for some constant }a.

    What is the relation beween S(n) and F(n), the n^{th} Fibonacci number?

    Crank out the first few terms:

    . \begin{array}{|c||c|c|}n & S(n) & F(n)\\ \hline<br />
1 & 1 & 1\\ <br />
2 & 1 & 1\\<br />
3 & 2+a & 2 \\<br />
4 & 3+2a & 3\\<br />
5 & 5+4a & 5 \\<br />
6 & 8 + 7a & 8 \\<br />
7 & 13 + 12a & 13 \\<br />
8 & 21 + 20a & 21 \\<br />
\vdots & \vdots & \vdots<br />
\end{array}


    And we see that: . S(n) \;=\;F(n) + [F(n)-1]\cdot a

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  6. #6
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    Thanks for all the posts
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