Nth term Fibonacci formula.

Question

I'm studying power series and in the video the professor showing an example of finding formula for n term of Fibonacci sequence.

at the middle of the video (6:00) https://www.youtube.com/watch?v=CR-nmp97Ayo

the professor rewrite this equation:

$$f(x)=\frac{x}{1-x-x^2}$$

to be:

$$f(x)=\frac{\frac{1}{\sqrt5}}{1-(x*\phi)}+\frac{-\frac{1}{\sqrt5}}{1-(x*(1-\phi))}$$

I try using quadratic formula what I get is this:

$$f(x)=\frac{x}{(x+\frac{1+\sqrt5}{2})(-x+\frac{-1+\sqrt5}{2})}$$

how can I go from my result of using quadratic formula to the result the professor show?

please give me very detail step by step.

Answer 1

$$\frac A{ax+1}+\frac B{bx+1}=\frac{(Ab+Ba)x+(A+B)}{(x+a)(x+b)}$$ and you are in a case such that

$$A+B=0.$$

Answer 2

This is partial fraction decomposition.

If you have $$\frac x {(x-a)(x-b)}=\frac A {x-a}+\frac B {x-b}$$ that is to say $$x=A(x-b)+B(x-a)=(A+B) x-(Ab+Ba)$$ Comparing terms, you then have equations $$1=A+B$$ $$0=Ab+Ba$$ Solve them for $A,B$ to get $$\frac x {(x-a)(x-b)}=\frac{a}{(a-b) (x-a)}-\frac{b}{(a-b) (x-b)}$$