数列题 a1=a2=1,a3=2,an+1=(3+anan-1)/(an-2) (n≥3),求an
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数列题 a1=a2=1,a3=2,an+1=(3+anan-1)/(an-2) (n≥3),求an
如图所示:
![](http://img.wesiedu.com/upload/b/92/b922ee9cc0524d56c6ceb0a2ad0eca8b.jpg)
如图所示:
![](http://img.wesiedu.com/upload/b/92/b922ee9cc0524d56c6ceb0a2ad0eca8b.jpg)
设F1=F2=1,F=F+Fn,则
Fn={[(1+√5)/2]^n-[(1-√5)/2]^2}/√5,
a4=(3+a3a2)/a1=5=F5,
a5=(3+a4a3)/a2=13=F7,
a6=(3+a5a4)/a3=34=F9,
a7=(3+a6a5)/a4=89=F11,
a8=(3+a7a6)/a5=233=F13,
猜n>3时an=F,①
以n+2代n,原式变为aan-aa=3,②
把①代入②左,FF-FF
={[(1+√5)/2]^(2n+3)-[(1-√5)/2]^(2n+3)}/√5*{[(1+√5)/2]^(2n-3)-[(1-√5)/2]^(2n-3)}/√5
-{[(1+√5)/2]^(2n+1)-[(1-√5)/2]^(2n+1)}/√5*{[(1+√5)/2]^(2n-1)-[(1-√5)/2]^(2n-1)}/√5
=(1/5){-[(1+√5)/2]^6-[(1-√5)/2]^6+[(1+√5)/2]^2+[(1-√5)/2]^2}(-1)
=(1/5)(-18+3)(-1)=3.猜想成立.
∴n>3时an=F={[(1+√5)/2]^(2n-3)-[(1-√5)/2]^(2n-3)}/√5.
Fn={[(1+√5)/2]^n-[(1-√5)/2]^2}/√5,
a4=(3+a3a2)/a1=5=F5,
a5=(3+a4a3)/a2=13=F7,
a6=(3+a5a4)/a3=34=F9,
a7=(3+a6a5)/a4=89=F11,
a8=(3+a7a6)/a5=233=F13,
猜n>3时an=F,①
以n+2代n,原式变为aan-aa=3,②
把①代入②左,FF-FF
={[(1+√5)/2]^(2n+3)-[(1-√5)/2]^(2n+3)}/√5*{[(1+√5)/2]^(2n-3)-[(1-√5)/2]^(2n-3)}/√5
-{[(1+√5)/2]^(2n+1)-[(1-√5)/2]^(2n+1)}/√5*{[(1+√5)/2]^(2n-1)-[(1-√5)/2]^(2n-1)}/√5
=(1/5){-[(1+√5)/2]^6-[(1-√5)/2]^6+[(1+√5)/2]^2+[(1-√5)/2]^2}(-1)
=(1/5)(-18+3)(-1)=3.猜想成立.
∴n>3时an=F={[(1+√5)/2]^(2n-3)-[(1-√5)/2]^(2n-3)}/√5.
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