meixi123
春芽
共回答了14个问题采纳率:92.9% 举报
(1)当n=1时,(a-1)a1=a(a1-1)得a1=a>0.
∵(a-1)Sn=a(an-1),
∴当n≥2时,(a-1)Sn-1=a(an-1-1),
两式相减得(a-1)an=a(an-an-1),化为an=aan-1.
∴an>0恒成立,且
an
an?1=a(n≥2),
∴{an}是等比数列.
又{an}的首项a1=a,公比为a,
∴an=an.
(2)当a=
1/2]时,由(1)得Sn=
1
2(1?
1
2n)
1?
1
2=1?
1
2n,
∴bn=1?
1
2n+λn+
λ
2n=1+λn+
λ?1
2n,
要使{bn}为等差数列,则b1+b3=2b2,
即1+λ+
λ?1
2+1+3λ+
λ?1
23=2(1+2λ+
λ?1
22),
解得λ=1,
又当λ=1时,bn=n+1,
∴{bn}为等差数列,
综上所述:λ=1.
(3)若a=1,则A={1},Sn=n,∴S2?A,不合题意;
若a>1,则A=[1,a],S2=a+a2>a,∴S2?A,不合题意;
若0<a<1,则A=[a,1],Sn=a+a2+…+an=
a(1?an)
1?a=[a/1?a?
a1+n
1?a].
∴Sn∈[a,
a
1?a).
要使Sn∈A,则
0<a<1
a
1?a≤1,解得,0<a≤
1
2.
综上所述,满足条件的正数a存在,a的取值范围为(0,
1
2].
1年前
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