Prove $1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$ $n \ge 1$

$F_n$; Fibonacci numbers

$\frac{1+\sqrt5}{2}=\phi$

Prove

$$1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$$

I can't go any further can anyone help please?



Expand the equation

Let 

$A=\phi^{4n+1}F_{2n}F_{2n+1}$

$B=\phi^{2n+1}F_{2n}$

$C=\frac{F_{2n+1}}{\phi{F_{2n}}}$

$$\left(1+\frac{1}{\sqrt{A}+B}\right)^2=C$$

$$1+\frac{2}{\sqrt{A}+B}+\frac{1}{(\sqrt{A}+B )^2}=C$$

This method doesn't seem to be working.

Can A, B and C be simplify more further?

I try and rationalise the denominator

$$\frac{2(\sqrt{A}-B)(A-B)+(\sqrt{A}-B)^2}{(A-B^2)^2}=C$$

Still nothing is been simplified.
notice that $A=B^2C$, and since $B&gt;0$, $\sqrt A = B\sqrt C$.

Since $\sqrt A + B &gt; 0$, $1 + 1/(\sqrt A + B) = \sqrt C \iff \sqrt A + B + 1 = \sqrt C(\sqrt A + B)$.

But $\sqrt A+B+1 = B\sqrt C + B + 1$ and $\sqrt C(\sqrt A+B) = BC + B\sqrt C$, so $ \sqrt A + B + 1 = \sqrt C(\sqrt A + B) \iff B+1 = BC \iff \phi^{2n+1}F_{2n}+1 = \phi^{2n}F_{2n+1}$

Now if this is true you should be able to prove it by induction or by using a formula for $F_n$.
Try simplifying using these 》
F2n+1= Fn+1^2 + Fn^2

F2n= Fn (Fn+1 + F2n+1 )

&amp; Then a stage will come when everything is in terms of Fn, Fn+1 &amp; Fn-1

Then use this formula

Fn+1 = Fn + Fn-1

I didn't simplify after this, but I guess it might help.