设{an}满足a1=3,an+1=1/2(an+1) 是否存在k,使数列{an-k}成等比数列,若存在,求出k的值.若不存在,说明理由.(2)若要使an《=1025/1024成立,求n取值范围

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设{an}满足a1=3,an+1=1/2(an+1) 是否存在k,使数列{an-k}成等比数列,若存在,求出k的值.若不存在,说明理由.(2)若要使an《=1025/1024成立,求n取值范围
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设{an}满足a1=3,an+1=1/2(an+1) 是否存在k,使数列{an-k}成等比数列,若存在,求出k的值.若不存在,说明理由.(2)若要使an《=1025/1024成立,求n取值范围
设{an}满足a1=3,an+1=1/2(an+1) 是否存在k,使数列{an-k}成等比数列,若存在,求出k的值.
若不存在,说明理由.(2)若要使an《=1025/1024成立,求n取值范围

设{an}满足a1=3,an+1=1/2(an+1) 是否存在k,使数列{an-k}成等比数列,若存在,求出k的值.若不存在,说明理由.(2)若要使an《=1025/1024成立,求n取值范围
(1)
∵a(n+1)=1/2(an+1)=1/2an+1/2
若数列{an-k}成等比数列,设公比为q
则[a(n+1)-k]=q(an-k)
∴a(n+1)=qan+k-qk
与a(n+1)=1/2an+1/2对比
q=1/2, k-qk=1/2,k=1
∴存在k=1使得
∴{an-1}是等比数列,公比为1/2
首项a1-1=3-1=2
∴an-1=2*(1/2)^(n-1)=2^(2-n)
∴an=1+2^(2-n)
(2)
an≤1025/1024
即1+2^(2-n)≤1+1/1024
∴2^(2-n)≤2^(-10)
∴2-n≤-10
∴n≥12
即n取值范围是n≥12,n∈N*

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