iccanobiF Sequence

Starting with F 1 = 1 F_1 = 1 , F 2 = 1 F_2 = 1 and then performing F n = F n 1 + F n 2 F_n = F_{n-1} + F_{n-2} , we get the sequence of Fibonacci Numbers , F n F_n . However, we can also extend this backwards, and look at when n < 1 n < 1 .

Using this, what is the value of F 2018 + F 2018 F_{2018} + F_{-2018} ?


The answer is 0.

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1 solution

Jacopo Piccione
Jun 21, 2018

First we notice that F x = { F x if x is odd F x if x is even F_{-x} = \begin{cases} \ F_{x} & \text{if x is odd } \\ - F_{x} & \text{if x is even } \end{cases} with x > 0 x>0 Since 2018 is even, F 2018 + F 2018 = F 2018 + F 2018 = 0 F_{-2018}+F_{2018}=-F_{2018}+F_{2018}=0

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