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Prove that for all $n \geq 1$, $F_{-n}$ = $(-1)^{n+1}F_n$ where F is the Fibonacci numbers.

I've already shown that the formula holds for $n = 1$ and $n = 2$. So I supposed the formula holds for $n$ and started with $F_{-(n+1)}$, and I am trying to get it to equal $(-1)^{n+1 + 1}F_{n+1}$.

Here are the steps I've taken so far:

$F_{-(n+1)}$ = $F_{-n+1} - F_{-n}$

Since the formula holds for $n$ I can change this to

$F_{-n+1} - (-1)^{n+1}F_n$

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

= $F_{-n+1} + (-1)^{n+2}(F_{n+1} - F_{n-1})$

= $F_{-n+1} + (-1)^{n+2}F_{n+1} + (-1)^{n+2}F_{n-1}$

So now I have the part I want in the middle, but with a couple extraneous pieces. Am I close to getting the desired result? Or is there another way I'm supposed to do this problem?

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    $\begingroup$ What is $\;F_{-n}\;$ ? I though that was a numbering index. Thanks. $\endgroup$ – DonAntonio Feb 29 '16 at 9:07
  • $\begingroup$ $F_n$ is Fibonacci numbers, and $F_{-n}$ is basically the same thing, except any positive values plugged in will yield the negative Fibonacci number. $\endgroup$ – Chris Feb 29 '16 at 9:17
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Extending the indices, we get $$ F_{-n}=F_{-n-1}+F_{-n-2} $$ which is the same as $$ (-1)^{n+3}F_{-(n+2)}=(-1)^{n+2}F_{-(n+1)}+(-1)^{n+1}F_{-n} $$ Therefore, $(-1)^{n+1}F_{-n}$ obeys the Fibonacci recursion. All we need to check is that $(-1)^1F_0=0$ and $(-1)^2F_{-1}=1$, and we have shown that $$ (-1)^{n+1}F_{-n}=F_n $$

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For even indices, what you are trying to prove -- i.e., $F_{-n}=-F_n$ -- IS true.
For odd indices, $F_{-n}=F_n$

\begin{array}{l|rrrrrrrrrrrrrrr} n&-7&-6&-5&-4&-3&-2&-1&0&1&2&3&4&5&6&7\\ F_n&13&-8&5&-3&2&-1&1&0&1&1&2&3&5&8&13 \end{array}

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