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已知数列{an}的前n项和为Sn,且满足an=2Sn+1(n∈N*).
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)若bn=(2n-1)•an,求数列{bn}的前n项和Tn.
答案:
(Ⅰ)当n=1时,a1=2S1+1=2a1+1,解得a1=-1.
当n≥2时,an=2Sn+1,an-1=2Sn-1+1,两式相减得an-an-1=2an,化简得an=-an-1,
所以数列{an}是首项为-1,公比为-1的等比数列,
可得an=(-1)n.
(Ⅱ)由(Ⅰ)得bn=(2n-1)•(-1)n,
当n为偶数时,bn-1+bn=2,Tn=n/2×2=n;
当n为奇数时,n+1为偶数,Tn=Tn+1-bn+1=(n+1)-(2n+1)=-n.
所以数列{bn}的前n项和Tn=(-1)n•n.
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