フィボナッチ数列

f(x)=(x)/(1-x-x^2)=(1/√5)/(1-αx)-(1/√5)/(1-βx)

α=(1+√5)/2、β=(1-√5)/2

an=1/√5・{α^n-β^n}

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間引いたフィボナッチ数列{F2n+1}を求めるために

{F2n+1}=1/√5・{α^2n+1-β^2n+1}

{Fn}^2=1/5・{α^n-β^n}^2

{Fn+1}^2=1/5・{α^n+1-β^n+1}^2

{Fn}^2+{Fn+1}^2=1/5・{α^2n+β^2n-2(αβ)^n+α^2n+2+β^2n+2-2(αβ)^n+1｝

==1/5・{α^2n+β^2n+α^2n+2+β^2n+2

==1/5・{α^2n+1(α+1/α)+β^2n+1(β+1/β)}

==1/5・{α^2n+1(√5)+β^2n+1(-√5)}｝

=1/√5・{α^2n+1-β^2n+1}

={F2n+1}

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{Fn+1}^2=1/5・{α^n+1-β^n+1}^2=1/5・{α・α^n-β・β^n}^2

=1/5・{α^2・α^2n+β^2・β^2n-2αβ・α^n-・β^n}

{Fn}^2=1/5・{α^n-β^n}^2

=1/5・{α^2n+β^2n-2・α^n・β^n}

{Fn}^2+{Fn+1}^2=1/5・{(α^2+1)α^2n+(β^2+1)β^2n-2(αβ+1)α^n・β^n｝

４で割って１または２余ることを示すのには至らない

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