Equation Involving Fibonacci Numbers

Question

I'm tasked with proving the following equivalency:

Let $F_n$ be the Fibonacci number at n, or $F_n = F_{n-1}+F_{n-2}, n >=2$

Show that:

$$\frac{1}{F_nF_{n+2}}=\frac{1}{F_nF_{n+1}}-\frac{1}{F_{n+1}F_{n+2}}$$

I tried filling in the definition $F_n$ which yielded the following:

$$\frac{1}{(F_{n-1}+F_{n-2})(F_{n+1}+F_n)}=\frac{1}{(F_{n-1}+F_{n-2})(F_{n}+F_{n-1})}-\frac{1}{(F_{n}+F_{n-1})(F_{n+1}+F_{n})}$$

Which looked promising, with binomial terms that might potentially cancel each other out or reduce cleanly. But after almost an hour of algebraic gymnastics I just can't get both sides to look exactly like the other.

EDIT: Getting a common denominator is key, I'm just struggling to find the exact form that causes everything to reduce in such a way that it proves the equivalence.

Answer 1

We have \begin{align*}\frac{1}{F_n F_{n+2}} + \frac{1}{F_{n+1} F_{n+2}} &= \frac{F_{n+1}F_{n+2} + F_{n}F_{n+2}}{F_{n}F_{n+1}F_{n+2}^2} \\ &= \frac{F_{n+1} + F_{n}}{F_{n}F_{n+1}F_{n+2}} \\ &= \frac{1}{F_{n}F_{n+1}}\end{align*} Thus $$\frac{1}{F_n F_{n+2}} = \frac{1}{F_{n}F_{n+1}}\ - \frac{1}{F_{n+1} F_{n+2}}$$

Answer 2

$$\begin{align} F_{n+2} &=F_{n+1}+F_{n}\\[2ex] F_{n+1} &=F_{n+2}-F_{n}\\[2ex] \frac{F_{n+1}}{F_nF_{n+1}F_{n+2}} &= \frac{F_{n+2}}{F_nF_{n+1}F_{n+2}}-\frac{F_{n}}{F_nF_{n+1}F_{n+2}}\\[2ex] \frac{1}{F_nF_{n+2}} &=\frac{1}{F_nF_{n+1}}-\frac{1}{F_{n+1}F_{n+2}} \end{align}$$