已知數(shù)列{a
n}的前n項和S
n滿足a
n+2S
nS
n-1="0" (n≥2),a
1=
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,求a
n.
a
n=
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.
∵當n≥2時,a
n=S
n-S
n-1,
∴S
n-S
n-1+2S
nS
n-1=0,
即
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-

=2, 4分
∴數(shù)列
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是公差為2的等差數(shù)列. 6分
又S
1=a
1=
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,∴
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=2,
∴
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=2+(n-1)·2=2n,
∴S
n=
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. 10分
∴當n≥2時,a
n=-2S
nS
n-1=-2·
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·
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=-
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, 12分
∴a
n=
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. 14分
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