# Fibonacci Identity Proof

I'm having a bit of trouble proving the following identity, where $F_n$ represents the $n$th Fibonacci number. The identity is $$\frac{1}{F_{2n}} = \frac{F_{n-1}}{F_n} - \frac{F_{2n-1}}{F_{2n}}$$ where $n$ is even and positive.

• Have you tried... induction? – Simply Beautiful Art Mar 5 '17 at 22:31
• Yes but since n has to be even that made things a bit more complicated – Andrew Ross Mar 5 '17 at 22:37
• although is ugly, maybe you can use the explicit formula for $F_{n}$ – Veridian Dynamics Mar 5 '17 at 22:53

## 1 Answer

We shall show the following formula by induction (The required equation is implied by this.) \begin{eqnarray*} F_{n-1}F_{2n}=(-1)^nF_n+F_nF_{2n-1} \end{eqnarray*} Observe that the recurrence relation can be rewritten as \begin{eqnarray*} F_{2n+2}=2F_{2n+1}-F_{2n}+F_{2n-2}. \end{eqnarray*} So now \begin{eqnarray*} F_{n}F_{2n+2}=(F_{n-1}+F_{n-2})(2F_{2n+1}-F_{2n}+F_{2n-2}) \end{eqnarray*} Using the inductive hypothesis on $F_{n-1}F_{2n}$ & $F_{n-2}F_{2n-2}$ \begin{eqnarray*} F_{n}F_{2n+2}=-(-1)^nF_n-F_nF_{2n-1}+(-1)^{n-1}F_{n-1}+F_{n-1}F_{2n-3}+2F_{n-1}F_{2n+1}+F_{n-1}F_{2n-2}+2F_{n-2}F_{2n+1}-F_{n-2}F_{2n} \end{eqnarray*} Now several applications of $F_n=F_{n-1}+F_{n-2}$ & the result follows.