a1=1a2=2an=a(n-1)+a(n-2)（n大于等于3)求an貌似是斐波那契数列

Question

a1=1 a2=2 an=a(n-1)+a(n-2)（n大于等于3) 求an 貌似是斐波那契数列

Answer 1

已知 An=An-1+An-2
设 An+k*An-1=p*(An-1+k*An-2) 构造等比数列
得
p-k=1
p*k=1
k1=(√5-1)/2 p1=(√5+1)/2
k2=(-√5-1)/2 p2=(-√5+1)/2
设Bn=An+1+k1*An
B1=A2+k1*A1=(√5+3)/2=[(√5+1)/2]^2
Bn=B1*p1^(n-1)=[(√5+1)/2]^(n+1)
=An+1+k1*An\x05①
设Tn=An+1+k2*An
T1=A2+k2*A1=(-√5+3)/2=[(-√5+1)/2]^2
Tn=T1*p2^(n-1)=[(-√5+1)/2]^(n+1)
=An+1+k2*An\x05②
由①②,
An=[ [(√5+1)/2]^(n+1) - [(-√5+1)/2]^(n+1)]/√5