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an equality on Fibonacci numbers

Source: 3-rd Hungary-Israel Binational Mathematical Competition 1992

May 24, 2007
algebra proposedalgebra

Problem Statement

We examine the following two sequences: The Fibonacci sequence: F0=0,F1=1,Fn=Fn−1+Fn−2F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 } for n≥2n \geq 2; The Lucas sequence: L0=2,L1=1,Ln=Ln−1+Ln−2L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2} for n≥2n \geq 2. It is known that for all n≥0n \geq 0 Fn=αn−βn5,Ln=αn+βn,F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, where α=1+52,β=1−52\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}. These formulae can be used without proof. Prove that ∑k=1n[αkFk+12]=F2n+1  ∀n>1.\sum_{k=1}^{n}[\alpha^{k}F_{k}+\frac{1}{2}]=F_{2n+1}\; \forall n>1.
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