已知数列{an}的前n项和为Sn,且满足an+2Sn×S(n-1)=0,a1=1/2.
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已知数列{an}的前n项和为Sn,且满足an+2Sn×S(n-1)=0,a1=1/2.
(1)求证:{1/Sn}是等差数列;
(2)求数列{an}的通项公式.
(1)求证:{1/Sn}是等差数列;
(2)求数列{an}的通项公式.
![已知数列{an}的前n项和为Sn,且满足an+2Sn×S(n-1)=0,a1=1/2.](/uploads/image/z/19730372-68-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%BB%A1%E8%B6%B3an%2B2Sn%C3%97S%28n-1%29%3D0%2Ca1%3D1%2F2.)
(1)an=Sn-S(n-1)
则Sn-S(n-1)+2Sn*S(n-1)=0
得Sn+Sn*S(n-1)=S(n-1)-Sn*S(n-1)
Sn(1+S(n-1)=S(n-1)(1-Sn)
(1-Sn)/Sn=(1+S(n-1)/S(n-1)
1/Sn-1=1/S(n-1)+1
得1/Sn-1/S(n-1)=2即{1/Sn}是公差为2的等差数列.
(2)由1/Sn-1/S(n-1)=2 得.S1=1/2
1/Sn-1/S1=2(n-1) 1/Sn=2(n-1)+2=2n
Sn=1/(2n)
an=Sn-S(n-1)=1/(2n)-1/(2n-2)
则Sn-S(n-1)+2Sn*S(n-1)=0
得Sn+Sn*S(n-1)=S(n-1)-Sn*S(n-1)
Sn(1+S(n-1)=S(n-1)(1-Sn)
(1-Sn)/Sn=(1+S(n-1)/S(n-1)
1/Sn-1=1/S(n-1)+1
得1/Sn-1/S(n-1)=2即{1/Sn}是公差为2的等差数列.
(2)由1/Sn-1/S(n-1)=2 得.S1=1/2
1/Sn-1/S1=2(n-1) 1/Sn=2(n-1)+2=2n
Sn=1/(2n)
an=Sn-S(n-1)=1/(2n)-1/(2n-2)
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