How to simplify the formula for $n$th Fibonacci number when $n=2$?

Question

When n is equal to 2 how do I simplify when the $n=2$ is put into the equation below (by the way I have to prove this formula by induction that when n= any number it will equal that number)

$$F_n=\frac 1{\sqrt 5} \left(\frac{1+\sqrt5}{2}\right)^n − \frac1{\sqrt 5} \left(\frac{1-\sqrt5}{2}\right)^n$$

This is what I did

When $n=2$ is added to the equation

$$\begin{align}F_n&=\frac 1{\sqrt 5} \left(\frac{1+\sqrt5}{2}\right)^2 − \frac1{\sqrt 5} \left(\frac{1-\sqrt5}{2}\right)^2\\ &=\frac{2+\sqrt 5}{2\sqrt 5} - \frac{2-\sqrt 5}{2\sqrt 5}\\ &=\frac{4\sqrt{5}}{2\sqrt 5}=2 \end{align}$$

I don't know if this is right or not so, it would be nice if someone helps me to simply this formula to be (equal 2)

Answer 1

In the first line, $(1+\sqrt 5)^2=1+2\sqrt 5 + 5=6+2\sqrt 5$ Similarly $(1-\sqrt 5)^2=6-2\sqrt 5$. Also the $2$'s in the denominators get squared. It should be $$\begin{align}F_2&=\frac 1{\sqrt 5} \left(\frac{1+\sqrt5}{2}\right)^2 − \frac1{\sqrt 5} \left(\frac{1-\sqrt5}{2}\right)^2\\ &=\frac{6+2\sqrt 5}{4\sqrt 5} - \frac{6-2\sqrt 5}{4\sqrt 5}\\ &=\frac{4\sqrt{5}}{4\sqrt 5}=1 \end{align}$$ and in the usual definition of the subscripts $F_2=1$

Answer 2

If you factor out the $\frac{1}{\sqrt{5}}$ you are left with difference of two squares so you need sum and difference of the terms in the parentheses, that is $(1)(\sqrt{5})$ then multiplied by the original factor taken out leaves F[2] = 1.