Fibonacci general term: How do I mend this broken derivation? I am working on a derivation of the general term for the Fibonacci Sequence. Here's where I want to be:
$$F(n)=\frac{1}{\sqrt5}(\phi^n-\overline{\phi}^n)$$
Here, $F(n)$ is the $n^\text{th}$ term, $\phi = \frac{1}{2}(1+\sqrt5)$ and $\overline{\phi}=\frac{1}{2}(1-\sqrt5)$
My generating function and power series for $|x|\lt 1$:
$$f(x)=\frac{x}{1-x-x^2}=x+x^2+2x^3+3x^4+5x^5+\cdots$$
By factoring the quadratic and rewriting as partial fractions, I'm left with:
$$f(x)=\tfrac{1}{\sqrt5}\cdot x\left(\frac{1}{x+\phi}-\frac{1}{x+\overline{\phi}}\right)$$
I think I go off track with the binomial expansion. Here's what I have:
$$f(x)=\frac{1}{\sqrt5}\cdot x\left[(x+\phi)^{-1}-(x+\overline{\phi})^{-1}\right]=\frac{1}{\sqrt5}\cdot x \sum_{k=0}^{\infty}\binom{-1}{k} x^{-1-k}(\phi^k-\overline{\phi}^k)$$
The $x$ before the $\sum$ means the $x$ term is $x^{-k}$. Also $\binom{-1}{k}=(-1)^k$. This leaves us with:
$$f(x)=\frac{1}{\sqrt5}\left(-\frac{\phi-\overline{\phi}}{x}+\frac{\phi^2-\overline{\phi}^2}{x^2}-\frac{\phi^3-\overline{\phi}^3}{x^3}+\frac{\phi^4-\overline{\phi}^4}{x^4}-...\right)$$
This is almost right, but every other term is negative. My result:
$$F(n)=\frac{(-1)^n}{\sqrt5}\left(\phi^n-\overline{\phi}^n\right)$$
The intrusive $(-1)^n$ leads me to suspect my handling of the coefficient, $\binom{-1}{k}$.
I'd appreciate help troubleshooting this. Thanks.
 A: Your approach seems overly complicated to me since it is not necessary to use the binomial series.
If you must use a generating function approach, let $$f(z) = \sum_{n=0}^\infty F_n z^n.$$  Note that $F_0 = 0$ according to your formula.  Then since $$F_n = F_{n-1} + F_{n-2},$$ we have
$$\begin{align}
f(z) &= 0 z^0 + 1 z^1 + z^2 \sum_{n=0}^\infty F_{n+2} z^n \\
&= z + z \sum_{n=0}^\infty F_{n+1} z^{n+1} + z^2 \sum_{n=0}^\infty F_n z^n \\
&= z + z f(z) + z^2 f(z).
\end{align}$$
Then $$f(z) = \frac{z}{1-z-z^2},$$ as you obtained.  However, from here, we proceed differently.  We observe the factorization $$1 - z - z^2 = (1 - \phi z)(1 - \bar \phi z).$$  Therefore, $$f(z) = \frac{1}{\sqrt{5}} \left( \frac{1}{1 - \phi z} - \frac{1}{1 - \bar \phi z}\right).$$  Then we simply use the formula for a geometric series:  $$f(z) = \frac{1}{\sqrt{5}} \left( \sum_{n=0}^\infty (\phi z)^n - (\bar \phi z)^n \right).$$  Equating coefficients gives $$F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \bar \phi^n\right),$$ as claimed.

If you do not need to use generating functions, then the approach I would use is as follows.  Observe $\phi + \bar \phi = 1$, and $\phi \bar \phi = -1$.  Then consider the identity $$(a^n - b^n)(a + b) = (a^{n+1} - b^{n+1}) - (ab)(a^{n-1} - b^{n-1}),$$ which is true for any $a, b$.  Then substituting $a = \phi$, $b = \bar \phi$, we obtain $$(\phi^n - \bar \phi^n)(1) = (\phi^{n+1} - \bar \phi^{n+1}) - (-1)(\phi^{n-1} - \bar\phi^{n-1}).$$  If we let $G_n = \phi^n - \bar \phi^n$ and rearranging terms, we get $$G_{n+1} = G_n + G_{n-1}.$$  Since $G_0 = 0$ and $G_1 = \sqrt{5}$, it follows that $F_n = G_n/\sqrt{5}$ for all $n$ and we are done.
A: $\dfrac1{x+\phi}=\dfrac1{\phi\left(1+\dfrac x\phi\right)}=\dfrac1\phi\left(1-\dfrac x\phi + \dfrac{x^2}{\phi^2}-\dfrac{x^3}{\phi^3}+\cdots\right)$
so $f(x)=\dfrac x{\sqrt5}\left(\dfrac1{x+\phi}-\dfrac1{x+\overline\phi}\right)$
$=\dfrac x{\sqrt5}\left(\dfrac1\phi-\dfrac1{\overline{\phi}}\right)-\dfrac{x^2}{\sqrt5}\left(\dfrac1{\phi^2}-\dfrac1{\overline\phi^2}\right)+\dfrac{x^3}{\sqrt5}\left(\dfrac1{\phi^3}-\dfrac1{\overline\phi^3}\right)-\dfrac{x^4}{\sqrt5}\left(\dfrac1{\phi^4}-\dfrac1{\overline\phi^4}\right)\cdots$,
whence $F_n=(-1)^{n+1}\dfrac1{\sqrt5}\left(\dfrac1{\phi^n}-\dfrac1{\overline{\phi}^n}\right)$.
Also, $\dfrac1{\phi^n}=(-1)^n\overline{\phi}^n$, so $F_n=\dfrac1{\sqrt5}\left(\phi^n-\overline{\phi}^n\right)$.
