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,αβ=-1,α^2+β^2=3、β=-1/α

an=1/√5・{α^n-β^n}

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F2n+1=Fn^2+Fn+1^2l

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a2n+2=1/√5・{α^2n+2-β^2n+2}

a2n=1/√5・{α^2n-β^2n}

(a2n+2)^2=1/5・{α^4n+4-2(αβ)^2n+2+β^4n+4}=1/5・{α^4n+4-2+β^4n+4}

(a2n)^2=1/5・{α^4n-2(αβ)^2n+β^4n}=1/5・{α^4n-2+β^4n}

(ａ^2+ｂ^2-１）=1/5・{α^4n+4-2+β^4n+4+α^4n-2+β^4n-5}

(ａ^2+ｂ^2-１）=1/5・{α^4n+4+β^4n+4+α^4n+β^4n-9}

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a2n+2・a2n=1/5・{α^2n+2-β^2n+2}{α^2n-β^2n}

a2n+2・a2n=1/5・{α^4n+2+β^4n+2-α^2n+2β^2n-α^2nβ^2n+2}

a2n+2・a2n=1/5・{α^4n+2+β^4n+2-α^2(αβ)^2n-(αβ)^2nβ^2}

a2n+2・a2n=1/5・{α^4n+2+β^4n+2-α^2-β^2}

a2n+2・a2n=1/5・{α^4n+2+β^4n+2-3}

3a2n+2・a2n=1/5・{3α^4n+2+3β^4n+2-9}

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5(ａ^2+ｂ^2-１）-15ａｂ

={α^4n+4+β^4n+4+α^4n+β^4n-3α^4n+2-3β^4n+2}

={α^4n+2(α^2+α^-2-3)+β^4n+2(β^2+β^-2-3)}

={α^4n+2(α^2+β^2-3)+β^4n+2(α^2+β^2-3)}

＝0

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ａ＝Ｆ2k，ｂ＝Ｆ2k+2のとき、

(ａ^2+ｂ^2-１）／ａｂ＝3

が成り立つ

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