2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Have you tried... induction? $\endgroup$ – Simply Beautiful Art Mar 5 '17 at 22:31
  • 2
    $\begingroup$ Yes but since n has to be even that made things a bit more complicated $\endgroup$ – Andrew Ross Mar 5 '17 at 22:37
  • $\begingroup$ although is ugly, maybe you can use the explicit formula for $F_{n}$ $\endgroup$ – Veridian Dynamics Mar 5 '17 at 22:53
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.