已知数列{an}的前n项和为Sn,且an=Sn·Sn-1（n大于等于2,Sn不为0）,a1=2/9 求{an}的通项公式
人气：140 ℃ 时间：2019-12-14 17:44:39
解答
an=Sn·Sn-11=an/(Sn·Sn-1)=1/Sn-1 - 1/Sn1=1/Sn-2 - 1/Sn-1……………………1=1/S1 - 1/S2 n-1式相加有n-1=1/S1 - 1/Sn S1=a1所以 1/Sn=1/a1-(n-1)=11/2-n1/Sn-1=11/2-(n-1)=13/2-n所以 an=Sn-Sn-1=1/(11/2-n)-1/(...
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