题目内容
设数列{an}的前n项和为Sn,且Sn=(m+1)-man
对任意正整数n都成立,其中m为常数,且m<-1.
(1)求证:{an}是等比数列;
(2)设数列{an}的公比q=f(m),数列{bn}满足:b1=
a1,bn=f(bn-1)(n≥2,n∈N*). 试问当m为何值时,
成立?
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(1)求证:{an}是等比数列;
(2)设数列{an}的公比q=f(m),数列{bn}满足:b1=

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(1) 证明略,(2) 
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(1)由已知Sn+1=(m+1)-man+1 ①, Sn=(m+1)-man ②,
由①-②,得an+1=man-man+1,即(m+1)an+1=man对任意正整数n都成立.
∵m为常数,且m<-1
∴
,即{
}为等比数列.
(2)当n=1时,a1=m+1-ma1,∴a1=1,从而b1=
由(1)知q=f(m)=
,∴bn=f(bn-1)=
(n∈N*,且n≥2)
∴
,即
,
∴{
}为等差数列。 ∴
=3+(n-1)=n+2,
(n∈N*).
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.
由①-②,得an+1=man-man+1,即(m+1)an+1=man对任意正整数n都成立.
∵m为常数,且m<-1
∴
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(2)当n=1时,a1=m+1-ma1,∴a1=1,从而b1=
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由(1)知q=f(m)=
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∴
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∴{
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