已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=3,(n-1)Sn=nSn-1+n2-n(n≥2).
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)令bn=an2n,求數(shù)列{bn}的前n項(xiàng)和Tn.
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1
=
3
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1
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n
2
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n
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2
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b
n
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2
n
【考點(diǎn)】錯(cuò)位相減法.
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【解答】
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發(fā)布:2024/8/16 12:0:1組卷:10引用:1難度:0.6
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