已知數(shù)列{an}滿足an+1-an=2,其前8項(xiàng)的和為64;數(shù)列{bn}是公比大于0的等比數(shù)列,b1=3,b3-b2=18.
(1)求數(shù)列{an}和{bn}的通項(xiàng)公式;
(2)記cn=an+2-1anan+1bn,n∈N*,求數(shù)列{cn}的前n項(xiàng)和Tn;
(3)記dn=(-1)n+12?an,n為奇數(shù) an2+1bn2,n為偶數(shù)
,求S2n=2n∑k=1dk.
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( - 1 ) n + 1 2 ? a n , n 為奇數(shù) |
a n 2 + 1 b n 2 , n 為偶數(shù) |
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【考點(diǎn)】錯(cuò)位相減法.
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發(fā)布:2024/6/27 10:35:59組卷:883引用:1難度:0.3
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