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    I have a formula in the form f(n) = ax^n + by^n
    And I want to know how I find the common difference/ common ratio between each term?

    I have worked out what both a,b,x and y are, if that helps!
    a = 1/sqrt(3) , b = -1/sqrt(3)
    x = (1+sqrt(3))/2 , y = (1-sqrt(3))/2

    Also, if this helps too:
    F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2

    I started by putting:
    F(n) = ax^n + by^n
    =1/(sqrt(3) [x^n - y^n]
    And therefore
    F(n+1) = ax^n+1 + by^n+1
    =1/(sqrt(3) [x^n+1 - y^n+1]

    But I'm not sure how to figure out the common difference/ common ratio, since x gets x times bigger, and y gets y times bigger, each term. But I can't find a value that does all in one go...

    I'm being askedto evaluate the sum to infinity of f(n)/2^n+1 so my first attempt is to try and find how much bigger f(n) gets each term, but I can't xD
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    (Original post by ComputerMaths97)
    I
    But I'm not sure how to figure out the common difference/ common ratio, since x gets x times bigger, and y gets y times bigger, each term. But I can't find a value that does all in one go...

    I'm being askedto evaluate the sum to infinity of f(n)/2^n+1 so my first attempt is to try and find how much bigger f(n) gets each term, but I can't xD
    I would forget about looking for a common difference or ratio, as there isn't one for the whole term.

    It may help to split the sequence into two.

    f(n) = ax^n + by^n

    Let g(n) = ax^n
    Let h(n) = by^n

    Both nice geometric series.

    And f(n) = g(n) + h(n)

    ...
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    Also: Why not post the question? Context is always useful.

    And BODMAS (or whatever precedence mnemonic you prefer) is not just an empty acronym. About 50% of what you've written is clearly wrong because of precedence rules. (E.g. you mean F(n+1) = ax^{n+1} + by^{n+1} but what you've actually written says F(n+1) = ax^n + 1 + b y^n+1).

    We can guess what you mean from context but those guesses are never 100% certain and if I have to make ten 90%-likely guesses about what you mean the likelihood is I'll be wrong at some point.

    In particular, I really don't know if you are trying to find \displaystyle \sum \dfrac{F(n)}{2^n+1} or \displaystyle \sum \dfrac{F(n)}{2^{n+1}}
 
 
 
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