I want to prove Binet's formula $F_{k} = \dfrac{1}{\sqrt{5}}\times\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^k-\left(\dfrac{1-\sqrt{5}}{2}\right)^k\right]$ for the $k_{th}$ fibonacci number.
I did as follows -
$F_{0}=0, F_{1} = 1$ .....
$F_{k} = F_{k-1}+F_{k-2}$
$F_kx^k=xF_{k-1}x^{k-1} + x^2F_{k-2}x^{k-2}$
summing over $k$ from $2$ to $\infty$,
$g(x)-1 = x(g(x)-1)+x^2g(x)\implies g(x) = \dfrac{x}{1-x-x^2} = \dfrac{-x}{x^2+x-1}$ where $g$ is the generating function of $F_0,F_1,F_2....$
Factoring we get
$g(x) = \dfrac{-x}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)} =-\left[\dfrac{\dfrac{\sqrt{5}-1}{2\sqrt{5}}\left(x-\dfrac{1+\sqrt{5}}{2}\right)+\dfrac{\sqrt{5}+1}{2\sqrt{5}}\left(x-\dfrac{1-\sqrt{5}}{2}\right)}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)}\right]$
$\implies g(x) = -\left[\dfrac{\sqrt{5}-1}{2\sqrt{5}}\cdot\dfrac{1}{\left(x-\frac{1-\sqrt{5}}{2}\right)} + \dfrac{\sqrt{5}+1}{2\sqrt{5}}\cdot\dfrac{1}{\left(x-\frac{1+\sqrt{5}}{2}\right)}\right] = \left[\dfrac{\sqrt{5}-1}{2\sqrt{5}}\cdot\dfrac{1}{\left(\frac{1-\sqrt{5}}{2}-x\right)} + \dfrac{\sqrt{5}+1}{2\sqrt{5}}\cdot\dfrac{1}{\left(\frac{1+\sqrt{5}}{2}-x\right)}\right]$
$\implies g(x) = \dfrac{\sqrt{5}-1}{2\sqrt{5}}\cdot\dfrac{\frac{2}{1-\sqrt{5}}}{\left(1-\frac{2x}{1-\sqrt{5}}\right)}+\dfrac{\sqrt{5}+1}{2\sqrt{5}}\cdot\dfrac{\frac{2}{1+\sqrt{5}}}{\left(1-\frac{2x}{1+\sqrt{5}}\right)} = \dfrac{1}{\sqrt{5}} \left[\dfrac{1}{1-\frac{2x}{1+\sqrt{5}}}-\dfrac{1}{1-\frac{2x}{1-\sqrt{5}}} \right]$
$\implies g(x) = \dfrac{1}{\sqrt{5}}\left[\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{2x}{1+\sqrt{5}}\right)^k-\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{2x}{1-\sqrt{5}}\right)^k\right] = \dfrac{1}{\sqrt{5}}\displaystyle\sum_{k=0}^{\infty}\left[\left(\dfrac{2}{1+\sqrt{5}}\right)^k-\left(\dfrac{2}{1-\sqrt{5}}\right)^k\right]x^k$
$\\$
So I'm getting $F_k = \dfrac{1}{\sqrt{5}}\times \left[\left(\dfrac{2}{1+\sqrt{5}}\right)^k-\left(\dfrac{2}{1-\sqrt{5}}\right)^k\right]$ wheres actually $F_{k} = \dfrac{1}{\sqrt{5}}\times\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^k-\left(\dfrac{1-\sqrt{5}}{2}\right)^k\right]$.
Where did I go wrong? I cant figure it out. Any help is very appreciated. Thanks!
Since the roots of $x^2+x-1$ are $\frac{-1\pm\sqrt{5}}{2}$, at the begining it should be $$g(x) = \dfrac{-x}{\left(x+\frac{1+\sqrt{5}}{2}\right)\left(x+\frac{1-\sqrt{5}}{2}\right)}.$$ and therefore, at the end, we find $$F_k = -\dfrac{1}{\sqrt{5}}\times \left[\left(\dfrac{-2}{1+\sqrt{5}}\right)^k-\left(\dfrac{-2}{1-\sqrt{5}}\right)^k\right].$$ We recover the classic Binet's formula, by noting that, $$\frac{-2}{1+\sqrt{5}}=\frac{-2}{1+\sqrt{5}}\cdot\frac{1-\sqrt{5}}{1-\sqrt{5}}= \frac{-2(1-\sqrt{5})}{1-5}=\frac{1-\sqrt{5}}{2}$$ and $$\frac{-2}{1-\sqrt{5}}=\frac{-2}{1-\sqrt{5}}\cdot\frac{1+\sqrt{5}}{1+\sqrt{5}}= \frac{-2(1+\sqrt{5})}{1-5}=\frac{1+\sqrt{5}}{2}.$$