I have been given $f(x)=∑_1^∞a_n⋅x^n$, where $(a_n)$ is the Fibonacci Sequence starting at $a_1=a_2=1$. So far, I have shown that this series is convergent for $|x|<1/2$, and that $(x^2+x−1)⋅f(x)=−x.$
Now, using the partial fraction decomposition of $x/(x^2+x-1)$, which I got to be $(1+1/\sqrt5)/(2x+1+\sqrt5) + (1-1/\sqrt5)/(2x+1-\sqrt5)$ and using the power series of $1/(x-a)$, I must obtain another power series for $f$.
I am not confident in my answer for the partial fraction decomposition (even though it multiplies out into the correct fraction), as clearly using the power series for $1/(x+a)$ makes more sense. Regardless, I converted my decomposition into the following pair of power series:
$(1+1/\sqrt5)/(2x+1+\sqrt5) = (1/\sqrt5)\sum_{k=0}^\infty [(-2x)/(1+\sqrt5)]^k$
$(1-1/\sqrt5)/(2x+1-\sqrt5)=(-1/\sqrt5)\sum_{k=0}^\infty [(-2x)/(1-\sqrt5)]^k$
So my final answer for $f(x)$ ought to be the sum of these two times $-1$, but I have clearly messed up at some stage, because that left me with $f(x)=(1/\sqrt5)\sum_0^\infty(-\sqrt5)^kx^k$ which I doubt is correct.
I'd appreciate if someone could point out where I went wrong. Hopefully I at least did part of this question correctly.
The final part of this question, which I've yet to attempt, asks: "Using the fact that a function can have only one representation as a power series centred at zero show that $a_n =([(1+√5)/2]^n − [(1−√5)/2]^n)/2$"
Could I please be given a hint on the first steps to take with this question? What is it I am supposed to do with my answer for $f$, once I have the correct power series?
