log1/2(Sn)=n+1/2则Sn=(1/2)^(n+1/2)=2^(-n-1/2) S(n-1)=2^(-n+1-1/2)=2^(-n+1/2)∴an=Sn - S(n-1) =2^(-n-1/2) - 2^(-n+1/2) =2^(-n-1/2)[1-2^(-n+1/2+n+1/2)] = -2^(-n-1/2)
已知数列an的前n项和Sn满足log1/2,Sn=n+1/2,求an
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