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Thread: Fibonacci sequence

  1. #1
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    Fibonacci sequence

    Would the formulas

    $\displaystyle F_{n+3} + F_{n} = 2F_{n+2}$ and $\displaystyle F_{n+3} - F_{n} = 2F_{n+1}$ for n=0,1,2,...

    necessarily remain true if the sequence
    Fn was replaced by a sequence with the same recurrence relation as the Fibonacci sequence but different initial terms? Justify your answer briefly.

    I'm not sure, but would the formula remain true, but with different initial terms the sequence would have different numbers? I'm not quite sure how to tackle this. Thanks for any pointers.
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  2. #2
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    I think those can be deduced from the basic recurrence relation $\displaystyle F_{n+2}=F_{n+1}+F_n$. If you start substituting on the left-hand side, the right-hand side comes out pretty quickly in both cases.

    Post again in this thread if you're still having trouble.
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  3. #3
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    Quote Originally Posted by hollywood View Post
    I think those can be deduced from the basic recurrence relation $\displaystyle F_{n+2}=F_{n+1}+F_n$. If you start substituting on the left-hand side, the right-hand side comes out pretty quickly in both cases.

    Post again in this thread if you're still having trouble.
    Hi, me again! Sorry, but I am still unsure of the answer.

    I know from earlier workings that $\displaystyle F_{n}=F_{n+2}-F_{n+1}$ and $\displaystyle F_{n+3}=F_{n+2}+F_{n+1}$ but I am not sure how to answer the question.

    Any help is very much appreciated
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  4. #4
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    Quote Originally Posted by cozza View Post
    Would the formulas

    $\displaystyle F_{n+3} + F_{n} = 2F_{n+2}$ and $\displaystyle F_{n+3} - F_{n} = 2F_{n+1}$ for n=0,1,2,...

    necessarily remain true if the sequence
    Fn was replaced by a sequence with the same recurrence relation as the Fibonacci sequence but different initial terms? Justify your answer briefly.

    I'm not sure, but would the formula remain true, but with different initial terms the sequence would have different numbers? I'm not quite sure how to tackle this. Thanks for any pointers.
    Quote Originally Posted by cozza View Post
    Hi, me again! Sorry, but I am still unsure of the answer.

    I know from earlier workings that $\displaystyle F_{n}=F_{n+2}-F_{n+1}$ and $\displaystyle F_{n+3}=F_{n+2}+F_{n+1}$ but I am not sure how to answer the question.

    Any help is very much appreciated
    These are essentially the same formula. If you add $\displaystyle F_{n+1}$ to both sides of the first formula, you get $\displaystyle F_{n}+F_{n+1}=F_{n+2}-F_{n+1}+F_{n+1}=F_{n+2}$ so $\displaystyle F_{n+2}=F_{n+1}+F_{n}$, which is the second formula with n replaced by by n-1 (i.e. all three subscripts decreased by 1).

    But let's take your two formulas and do some algebra:

    $\displaystyle F_{n}=F_{n+2}-F_{n+1}$

    $\displaystyle F_{n+3}=F_{n+2}+F_{n+1}$

    adding $\displaystyle F_{n+1}$ to the first gives:

    $\displaystyle F_{n}+F_{n+1}=F_{n+2}$

    and substituting into the second gives:

    $\displaystyle F_{n+3}=F_{n}+F_{n+1}+F_{n+1}=F_{n}+2F_{n+1}$

    and subtracting $\displaystyle F_{n}$ from both sides gives:

    $\displaystyle F_{n+3}-F_{n}=2F_{n+1}$.

    That's the second formula you were looking for.

    Now, if you take:

    $\displaystyle F_{n+3}=F_{n+2}+F_{n+1}$

    and instead of substituting for $\displaystyle F_{n+2}$, you substitute for $\displaystyle F_{n+1}$ using $\displaystyle F_{n+1}=F_{n+2}-F_{n}$, you get:

    $\displaystyle F_{n+3}=F_{n+2}+F_{n+2}-F_{n}$

    $\displaystyle F_{n+3}+F_{n}=2F_{n+2}$, which is the first formula you were looking for.

    - Hollywood
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  5. #5
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    Quote Originally Posted by hollywood View Post
    These are essentially the same formula. If you add $\displaystyle F_{n+1}$ to both sides of the first formula, you get $\displaystyle F_{n}+F_{n+1}=F_{n+2}-F_{n+1}+F_{n+1}=F_{n+2}$ so $\displaystyle F_{n+2}=F_{n+1}+F_{n}$, which is the second formula with n replaced by by n-1 (i.e. all three subscripts decreased by 1).

    But let's take your two formulas and do some algebra:

    $\displaystyle F_{n}=F_{n+2}-F_{n+1}$

    $\displaystyle F_{n+3}=F_{n+2}+F_{n+1}$

    adding $\displaystyle F_{n+1}$ to the first gives:

    $\displaystyle F_{n}+F_{n+1}=F_{n+2}$

    and substituting into the second gives:

    $\displaystyle F_{n+3}=F_{n}+F_{n+1}+F_{n+1}=F_{n}+2F_{n+1}$

    and subtracting $\displaystyle F_{n}$ from both sides gives:

    $\displaystyle F_{n+3}-F_{n}=2F_{n+1}$.

    That's the second formula you were looking for.

    Now, if you take:

    $\displaystyle F_{n+3}=F_{n+2}+F_{n+1}$

    and instead of substituting for $\displaystyle F_{n+2}$, you substitute for $\displaystyle F_{n+1}$ using $\displaystyle F_{n+1}=F_{n+2}-F_{n}$, you get:

    $\displaystyle F_{n+3}=F_{n+2}+F_{n+2}-F_{n}$

    $\displaystyle F_{n+3}+F_{n}=2F_{n+2}$, which is the first formula you were looking for.

    - Hollywood
    So presumably the formulas would remain true if the sequence Fn was replaced by a sequence with the same recurrence relation as the Fibonacci sequence but different initial terms, and that is proved by replacing the initial term

    $\displaystyle F_{n+2}=F_{n+1}+F_n$ with F_{n+3}+F_{n}=2F_{n+2}[/tex] and F_{n+3}-F_{n}=2F_{n+1}[/tex]
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  6. #6
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    Quote Originally Posted by cozza View Post
    So presumably the formulas would remain true if the sequence Fn was replaced by a sequence with the same recurrence relation as the Fibonacci sequence but different initial terms, and that is proved by replacing the initial term

    $\displaystyle F_{n+2}=F_{n+1}+F_n$ with F_{n+3}+F_{n}=2F_{n+2}[/tex] and F_{n+3}-F_{n}=2F_{n+1}[/tex]
    Yes, since the first is true for all $\displaystyle n\ge0$, the other two are true for all $\displaystyle n\ge0$.

    - Hollywood
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