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Wolfram Alpha gives the $100$th fibonacci number to be $354224848179261915075$ and the $104$th fibonacci number to be $2427893228399975082453$. Just from this, can we deduce what the $102$th fibonacci will be? Is it at all possible, and if it is possible, can we use it to predict the $n$th fibonacci number?

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Generalization:

$$F_{n+4} = F_{n+3} + F_{n+2} = (F_{n+2} + F_{n+1}) + F_{n+2} = 2\cdot F_{n+2} + F_{n+1}$$

Since

$$F_{n+2} = F_{n+1} + F_{n} \implies F_{n+1} = F_{n+2} - F_{n}$$

$$\therefore F_{n+4} = 2\cdot F_{n+2} + (F_{n+2} - F_{n}) = 3\cdot F_{n+2} - F_{n}$$

$$\implies F_{n+2} = \dfrac{F_{n+4} + F_{n}}{3}$$

Now substituting $n = 100$ we have:

$$F_{102} = \dfrac{F_{104} + F_{100}}{3} = \dfrac{2427893228399975082453 + 354224848179261915075}{3} = 927372692193078999176 $$

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    $\begingroup$ Essentially, any $F_n$ can be computed as a rational combination of $F_{n+i}$ and $F_{n+j}$ with coefficients that are functions of only $i$ and $j$. $\endgroup$ – marty cohen Nov 14 '15 at 23:13
  • $\begingroup$ @martycohen: Yes, that is true. $\endgroup$ – Yagna Patel Nov 14 '15 at 23:14

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