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So I'm given the generating function $F(x)={1+2x\over1-3x^2}$ I'm supposed to find the recurrence relation satisfied by fn. I managed to get it into 2 separate geometric series and derive $f_n = {5(3^n)-(-3^n)\over6}$ but can't derive it in terms of past values of $f_n$. Help please, I have exam tomorrow!!!

EDIT: I just realised the $f_n$ I derived is wrong, ignore that.

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$F(x)=\sum_{i\ge 0}a_ix^i$, where $a_0=1,a_1=2$ and $a_i=3a_{i-2}$ for $i\ge 2$.

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    $\begingroup$ Do you mind elaborating how you arrived at the answer? I have trouble expanding the function. $\endgroup$
    – huhehu
    Nov 10 '13 at 9:30
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    $\begingroup$ @huhehu: I multiplied the left hand of $(1-3x^2)\sum_{i\ge 0}a_ix^i=i+2x$ and equated coefficients. $\endgroup$ Nov 10 '13 at 9:48
  • $\begingroup$ Ah yes, I found it! Cheers! $\endgroup$
    – huhehu
    Nov 10 '13 at 9:57
  • $\begingroup$ @huhehu: It’s really just the reverse of the process that you use to derive a generating function from a recurrence. $\endgroup$ Nov 10 '13 at 9:57

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