fibonacci recurrence relation proof I've been trying to prove the closed form solution of fibonacci recurrence sequence and achieve this

$a_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^n−(\frac{1-\sqrt{5}}{2})^n]$

And so far I haven't achieve that, this is how I did it

$a_n=x(\frac{1+\sqrt{5}}{2})^n+y(\frac{1-\sqrt{5}}{2})^n$
 $a_0=0=x+y$
 $a_1=1=x(\frac{1+\sqrt{5}}{2})+y(\frac{1-√5}{2})$

thus, I was able to get $x=\frac{\sqrt{5}+5}{10}$ and $y=-\frac{\sqrt{5}+5}{10}$. Then plugging in $x$ and $y$ to the formula this is what I got


$a_n=\frac{\sqrt{5}+5}{10}(\frac{1+\sqrt{5}}{2})^n+(-\frac{\sqrt{5}+5}{10})(\frac{1-\sqrt{5}}{2})^n$

beyond that, I just can't prove the closed form from above, I'm stuck to this, since I can't or don't know how to further reduce $\frac{\sqrt{5}+5}{10}$.

Did I miss anything? or got something wrong?

From $x+y=0$ we have $y=-x$ and substituting into the second we obtain $x(\frac{1+\sqrt{5}}{2})+(-x)(\frac{1-\sqrt{5}}{2})=x(\frac{1+\sqrt{5}-1+\sqrt{5}}{2})=x\sqrt{5}=1$ and thus $x=\frac{1}{\sqrt{5}}$ so $y=-\frac{1}{\sqrt{5}}$ as required.
Alternatively, you can use the generating function of the fibonacci sequence as @SeraPhim mentioned.
