Some Proofs About Fibonacci Numbers: Relation 2

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Relation 2

We can obtain below from the recurrence deﬁnition (1) of F_n. Again, assume n to be large enough to allow some iterations.

Fn = Fn −2+Fn −1 = (Fn−2+Fn −3)+Fn −2 = (Fn−2+Fn − 3+Fn −4)+Fn −3

In above, we start seeing that the second smallest ﬁbonacci term is occurring twice. We can pick one of the two occurrences of such term and expand it using its recurrence deﬁnition, and repeat as below:

Fn = (Fn−2 + Fn−3 + Fn−4)+ Fn −3 {expanding F } n−3 = (Fn−2 + Fn−3 + Fn−4 + Fn−5)+ Fn −4 {repeating as above} = (Fn−2 + Fn−3 + ...+ Fn−k+1 + Fn−k) + Fn−k+1 3 ≤ k ≤ n − 1 {with k = n − 1} = (Fn−2 + Fn−3 + ...+ F2 + F1 )+ F2

Since F₂ = 1, we can write below relation for all n > 2:

n∑−2 Fn = ( Fi)+ 1 i=1

(3)

Example: F₉ = 34 = (1 + 1 + 2 + 3 + 5 + 8 + 13) + 1

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