设数列{An}满足a1=2,An+1=入An+2^n,n属于N*,入为常数
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设数列{An}满足a1=2,An+1=入An+2^n,n属于N*,入为常数
(1)若A2=0,求A3的值;
(2)是否存在实数入,使得数列{An}为等差数列,若存在,求数列{An}的通项公式,若不存在,请说明理由;
(3)设入=1,Bn=4n-7/An,数列{Bn}的前n 项和为Sn,求满足Sn>0的最小自然数n的值.
(1)若A2=0,求A3的值;
(2)是否存在实数入,使得数列{An}为等差数列,若存在,求数列{An}的通项公式,若不存在,请说明理由;
(3)设入=1,Bn=4n-7/An,数列{Bn}的前n 项和为Sn,求满足Sn>0的最小自然数n的值.
(1)
A1 = 2,A2 = 0
所以 λ = (0-2^1) / 2 = -1
所以A3 = -1x0 + 2^2 = 4
(2)
可以直接从前三项考虑
A1 = 2,A2 = 2λ+2,A3=λA2+4=2λ^2+2λ+4
若这三项为等差,则有A3 - A2 = A2 - A1
即有λ^2-λ+1=0,而△ = 1-4 = -3
A1 = 2,A2 = 0
所以 λ = (0-2^1) / 2 = -1
所以A3 = -1x0 + 2^2 = 4
(2)
可以直接从前三项考虑
A1 = 2,A2 = 2λ+2,A3=λA2+4=2λ^2+2λ+4
若这三项为等差,则有A3 - A2 = A2 - A1
即有λ^2-λ+1=0,而△ = 1-4 = -3
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