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}

間引いたフィボナッチ数列{F2^n}、すなわち、１，１，３，２１，９８７，・・・

α=(1+√5)/2、β=(1-√5)/2

F2^n=1/√5・{α^2^n-β^2^n}

では

　　Σ1/F2^n=(7-√5)/2

が成り立つ

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F(2^n-1)L(2^n)=F(2^n+1-1)+1

ΠL2^i=F(2^n+1)

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P=(1+1/2)(1+1/13)(1+1/610)・・・=Π(1+1/F2^n+1-1)

=Π(1+F(2^n+1-1)/(F2^n+1-1)

=Π(F(2^n-1)L(2^n))/(F2^n+1-1)

=Π(F(2^n-1)/(F2^n+1-1)・F2(2^m+1)

=F(2^m+1)/F(2^m+1-1)→φ

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ΣFi/2^i+1=1/2=1/F3

ΣFi/3^i+1=1/3=1/F5

ΣFi/8^i+1=1/55=1/F10

ΣFi/10^i+1=1/89=1/F11

1+Fnが連続する2整数の積である場合は

2・3=1+F5

7・8=1+F10

1・2=1+F1

9・10=1+F11

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また、

ΣFi-1/10^i=1/89=1/F11

ΣFi-1/10^2i=1/9899

ΣFi-1/10^3i=1/998999

ΣFi-1/(-10^2i)=1/10099

ΣFi-1/(-10^3i)=1/1000999

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ΣFk/2^k=2

(証)

ΣFkx^k=x/(1-x-x^2)は|x|＜1/αのとき収束する.

x=1/2とおくとΣFk/2^k=2が得られる

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ΣFk+1/2^k=4

(証)

L=ΣFk+1/2^k=Σ(Fk+3-Fk+2)/2^k=2^3Σ(Fk+3/2^k+3)-2^2(Fk+2/2^k+2

=8Σ(Fk/2^k-F1/2-F2/4)-4Σ(Fk/2^k-F1/2)

=8(L2-1/2-1/4)-4(L/2-1/2)=2L-4

L=4

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Σ(k+1)^2/2^k・Fk=116

(証)

Σ(k+1)^2/2^k・Fkx^k=Σx(4-3x+5x^2)/(1-x-x^2)^3

X=1/2とおくと得られる

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Σ1/FnFn+2=1

Σ1/FnFn+4=7/18

Σ1/FnFn+2Fn+3=1/4

ΣFn/Fn+1Fn+2=1

ΣFn+1/FnFn+3=5/4

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Σ1/FnFn+2=Σ(1/FnFn+1-1/Fn+1Fn+2)=1/F1F2-1/Fn+1Fn+2→1

Σ1/FnFn+2Fn+3=Σ(1/FnFn+1Fn+2-1/Fn+1Fn+2Fn+3)=Σ(Fn+3-Fn)/FnFn+1Fn+2Fn+3

1/F1F2F3-1/Fn+1Fn+2Fn+3=2Σ(Fn+1)/FnFn+1Fn+2Fn+3→1/2

Σ1/FnFn+2Fn+3=1/4

ΣFn+1/FnFn+3=1/2Σ(1/Fn-1/Fn+3)

=1/2(1/F1+1/F2+1/F3-1/Fn+1-1/Fn+2-1/Fn+3)

=1/2(5/2-1/Fn+1-1/Fn+2-1/Fn+3)→5/4

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