$${x\over 1-x-x^2}$$
This is the closed form for series without coeff of $x^i$ being $F_i$ (ith Fibonacci Number). How can i expand the it without already knowing the series?
For example say we have $${1\over 1+x+x^2}$$ How do i expand it? (without taylor series because it might be difficult to find patterns this way)
On
For given, $a,b,c,d$ use Partial Fraction Decomposition
$$\dfrac{ax+b}{1+cx+dx^2}=\dfrac A{1-\alpha x}+\dfrac B{1-\beta x}$$
where $\alpha+\beta=c,\alpha\beta=d, \alpha\ne\beta$ as $\alpha=\beta\implies1+cx+dx^2$ a perfect square.
Now use Binomial series like $(1-\alpha x)^{-1}=\sum_{r=0}^\infty(\alpha x)^r$ which is readily available from Geometric Series
How to expand $${x\over 1-x-x^2}$$ in a power series?
Begin with $$x = (1-x-x^2) \sum_{i=0}^\infty a_i x^i$$ expand the right side, collect like terms, equate to the left side. You end up with a recurrence.
Try it! That way you will learn more than if we just do it for you.
After you do that, maybe try something more general. Any rational function has an expansion as a power series (perhaps a Laurent series) where the coefficients satisfy a linear recurrence with constant coefficients.
added It seems you tried it but it didn't work out.
Let's go: \begin{align} x &= (1-x-x^2) \sum_{i=0}^\infty a_i x^i \\ &= \sum_{i=0}^\infty a_i x^i - x\sum_{i=0}^\infty a_i x^i - x^2\sum_{i=0}^\infty a_i x^i \\ &= \sum_{i=0}^\infty a_i x^i - \sum_{i=0}^\infty a_i x^{i+1} - \sum_{i=0}^\infty a_i x^{i+2} \\ &= \sum_{i=0}^\infty a_i x^i - \sum_{i=1}^\infty a_{i-1} x^{i} - \sum_{i=2}^\infty a_{i-2} x^{i} \\ &= a_0 + (a_1-a_0)x +\sum_{i=2}^\infty (a_i - a_{i-1} - a_{i-2}) x^{i} \end{align} Equate terms on the two sides: \begin{align} 0 &= a_0 \\ 1 &= a_1-a_0 \\ 0 &= (a_i - a_{i-1} - a_{i-2})\qquad \text{for } i \ge 2 \end{align} So our power series coefficients satisfy \begin{align} a_0 &= 0 \\ a_1 &= 1\\ a_i &= a_{i-1}+a_{i-2}\qquad\text{for }i \ge 2 \end{align}