短时间内一定采纳,谢谢)已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.(1)求证:{1/Sn}是等差数列(2)求an的表达式.(1)里是怎样做到 1/(sn) -
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![短时间内一定采纳,谢谢)已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.(1)求证:{1/Sn}是等差数列(2)求an的表达式.(1)里是怎样做到 1/(sn) -](/uploads/image/z/4348622-38-2.jpg?t=%E7%9F%AD%E6%97%B6%E9%97%B4%E5%86%85%E4%B8%80%E5%AE%9A%E9%87%87%E7%BA%B3%2C%E8%B0%A2%E8%B0%A2%EF%BC%89%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7BAn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%E4%B8%94%E6%BB%A1%E8%B6%B3an%3D2S%28n-1%29Sn%28n%3E%3D2%29%2Ca1%3D1.%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7BAn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%E4%B8%94%E6%BB%A1%E8%B6%B3an%3D2S%28n-1%29Sn%28n%3E%3D2%29%2Ca1%3D1.%281%29%E6%B1%82%E8%AF%81%3A%7B1%2FSn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%282%29%E6%B1%82an%E7%9A%84%E8%A1%A8%E8%BE%BE%E5%BC%8F.%281%29%E9%87%8C%E6%98%AF%E6%80%8E%E6%A0%B7%E5%81%9A%E5%88%B0+1%2F%28sn%29+-)
短时间内一定采纳,谢谢)已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.(1)求证:{1/Sn}是等差数列(2)求an的表达式.(1)里是怎样做到 1/(sn) -
短时间内一定采纳,谢谢)已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.
已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.
(1)求证:{1/Sn}是等差数列
(2)求an的表达式.
(1)里是怎样做到 1/(sn) - 1/[s(n-1)] = -2的?
越详细越好,短时间内一定采纳.
短时间内一定采纳,谢谢)已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.已知数列{An}的前n项和为Sn且满足an=2S(n-1)Sn(n>=2),a1=1.(1)求证:{1/Sn}是等差数列(2)求an的表达式.(1)里是怎样做到 1/(sn) -
这是因为an=Sn-S(n-1)=2S(n-1)Sn
两边都除以S(n-1)Sn即得:1/(sn)-1/[s(n-1)]=-2
于是知{1/Sn}是以1/S1为首项,公差为-2的等差数列
∴1/Sn=1/S1+(-2)(n-1)=3-2n
∴Sn=1/(3-2n)
从而an=Sn-S(n-1)=1/(3-2n)-1/(5-2n)=2/[(3-2n)(5-2n)]
an=Sn-S(n-1)带进去,约分,两边同时除以S(n-1)Sn
得1/(sn) - 1/[s(n-1)] = -2
然后an的表达式按上述办法求出,先求出Sn
1.因为an=2S(n-1)Sn
an=Sn-S(n-1)
所以2S(n-1)Sn=Sn-S(n-1)
两边同时除以S(n-1)Sn
所以1/(sn) - 1/[s(n-1)] = -2
所以{1/Sn}是等差数列
2.因为1/(sn) - 1/[s(n-1)] = -2
所以1/S2-1/S1=-2
1/S3-1/S2=-2...
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1.因为an=2S(n-1)Sn
an=Sn-S(n-1)
所以2S(n-1)Sn=Sn-S(n-1)
两边同时除以S(n-1)Sn
所以1/(sn) - 1/[s(n-1)] = -2
所以{1/Sn}是等差数列
2.因为1/(sn) - 1/[s(n-1)] = -2
所以1/S2-1/S1=-2
1/S3-1/S2=-2
...
1/Sn-1/S(n-1)=-2
所有相加 1/Sn-1/S1=-2(n-1)
因为a1=1
所以S1=1
1/Sn=-2n+3
Sn=1/-2n+3
S(n-1)=1/-2n+5
an=Sn-S(n-1)
=2/(3-2n)(5-2n)
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