Fn+1=Fn+Fn-1
兩邊加kFn
Fn+1+kFn=(k+1)Fn+Fn-1
當(dāng)k!=1時
Fn+1+kFn=(k+1)(Fn+1/(k+1)Fn-1)
令
Yn=Fn+1+kFn
若
當(dāng)k=1/k+1,且F1=F2=1時
因為
Fn+1+kFn=1/k(Fn+kFn-1)
=>
Yn=1/kYn-1
所以
Yn為q=1/k=1(1/k+1)=k+1的等比數(shù)列
那么當(dāng)F1=F2=1時
Y1=F2+kF1=1+k*1=k+1=q
根據(jù)等比數(shù)列的通項公式
Yn=Y1q^(n-1)=q^n=(k+1)^n
因為k=1/k+1=>k^2+k-1=0
解為 k1=(-1+sqrt(5))/2
k2=(-1-sqrt(5))/2
將k1,k2代入
Yn=(k+1)^n
,和Yn=Fn+1+kFn
得到
Fn+1+(-1+sqrt(5))/2Fn=((1+sqrt(5))/2)^2
Fn+1+(-1+sqrt(5))/2Fn=((1-sqrt(5))/2)^2
兩式相減得
sqrt(5)Fn=((1+sqrt(5))/2)^2-((1-sqrt(5))/2)^2
Fn=(((1+sqrt(5))/2)^2-((1-sqrt(5))/2)^2)/sqrt(5)