I have an exercise which finds a formula for Fibonacci sequence from a recurrence relation $$\begin{matrix}u_1=u_2=1\\u_{n+2}=u_{n+1}+u_n \end{matrix}$$ I know there are numerous results on the Internet but mine has a little difference which leads to a wrong answer
$$A=\frac{1}{1-x-x^2}$$ $$A=\frac{1}{(2x+1-\sqrt{5})(2x+1+\sqrt{5})}$$ $$A=\frac{\alpha}{2x+1-\sqrt{5}}+\frac{\beta}{2x+1+\sqrt{5}}$$ We have $2\alpha x+\alpha+\sqrt{5}\alpha+2\beta x+\beta-\sqrt{5}\beta=1$
$=>\alpha=\frac{1}{\sqrt{5}}; \beta=-\frac{1}{\sqrt{5}}$ $$A=\frac{1}{\sqrt{5}}(\frac{1}{2x+1-\sqrt{5}}-\frac{1}{2x+1+\sqrt{5}})$$ $$A=\frac{1}{\sqrt{5}}(\frac{1}{1-\sqrt{5}}.\frac{1}{1+\frac{2}{1-\sqrt{5}}x}-\frac{1}{1+\sqrt{5}}.\frac{1}{1+\frac{2}{1+\sqrt{5}}x})$$ we have $\frac{1}{1+x}=1-x+x^2-x^3+...$ $$=>u_n=\frac{(-1)^{n-1}}{\sqrt{5}}[\frac{1}{1-\sqrt{5}}.(\frac{2}{1-\sqrt{5}})^{n-1}-\frac{1}{1+\sqrt{5}}.(\frac{2}{1+\sqrt{5}})^{n-1}]$$ The sequence base on $u_n$ is $u_1=-\frac{1}{2};u_2=-\frac{1}{2};u_3=-\frac{2}{2};u_4=-\frac{3}{2};u_5=-\frac{5}{2};u_6=-\frac{8}{2};u_7=-\frac{13}{2};...$
Where does the $-\frac{1}{2}$ come from?
Any help would be appreciated