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先求通项公式
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![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先求通项公式](/uploads/image/z/8786826-18-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2Can%3Da1%2B1%2F2a2%2B1%2F3a3%2B%E2%80%A6%2B1%2F%28n-1%29a%28n-1%29%2C%E8%8B%A5an%3D2006%2C%E5%88%99n%3D___%E4%B8%BA%E4%BB%80%E4%B9%88a1%3Da2%3D1%E5%85%88%E6%B1%82%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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先求通项公式
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
先求通项公式
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先求通项公式
an=a1+(1/2)a2+(1/3)a3+…+[1/(n-2)]a(n-2)+[1/(n-1)]a(n-1),
a(n-1)=a1+(1/2)a2+(1/3)a3+…+[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
n=2006
因为an=2006
则必须为整数。
因为a1=1,所以a2=1/(2-1)a(2-1)=1=a1
所以an=1+(n-2)/2
又因为an=2006
所以n=4012
a2=a1/1=1
a2=a1=1
n>=3时
an+1=a1+1/2a2+......+1/n-1an-1+1/nan
两式相减得an+1-an=1/nan
即an+1=n+1/nan
即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
2009