已知数列{an}的前n项和为Sn,且2Sn=2-(2n-1)an(n属于N*)(1)设bn=(2n+1)Sn,求数列{b
1个回答
n=1时2a1=2-a1,a1=2/3.
n>1时2Sn=2-(2n-1)(Sn-S),
∴(2n+1)Sn-[2(n-1)+1]S=2,
即bn-b=2,b1=3a1=2,
∴bn=2n.
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