<s>[INST] How to simplify the formula for $n$th Fibonacci number when $n=2$? 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)
 [/INST] 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 $F_2=1$
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