Question

已知数列{an}满足a1=1,an=a1+1/2a2+1/3a3+…+1/(n-1)a(n-1),若an=2006，则n=___

已知数列{an}满足a1=1,an=a1+1/2a2+1/3a3+…+1/(n-1)a(n-1),若an=2006，则n=___
为什么a1=a2=1
速求解
先求通项公式

Answer 1

an=a1+(1/2)a2+(1/顷段历3)a3+…雀搜+[1/(n-2)]a(n-2)+[1/(n-1)]a(n-1),
a(n-1)=a1+(1/2)a2+(1/3)a3+…燃友+[1/(n-2)]a(n-2)
an-a(n-1)=[1/(n-1)]a(n-1)
an=[n/(n-1)]a(n-1)
(1/n)an=[1/(n-1)]a(n-1)
(1/n)an=[1/(n-1)]a(n-1)=……=(1/2)a2=a1=1
an=2006时
(1/n)2006=1
n=2006

Answer 2

a2=a1/1=1
a2=a1=1
n>=3时
an+1=a1+1/2a2+......+1/n-1an-1+1/nan
两式颤碰相减茄羡谈得an+1-an=1/nan
即an+1=n+1/nan
即派敏an+1/an=n+1/n
an+1/a2=(an+1/an)(an/an-1)......(a3/a2)
=(n+1/n)(n/n-1)......(3/2)
=n+1/2
即an+1=n+1/2
即an=n/2
an=2006,
n=4012

Answer 3

n=2006
因为an=2006
则必须为整数。

追问

先求通项公式

Answer 4

因为a1=1,所猜辩以a2=1/(2-1)a(2-1)=1=a1
所以蔽团an=1+(n-2)/2
又因为an=2006
所以穗并缺n=4012

Answer 5

2009