已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2 (1)判断{1/Sn},、{An}是否是等差数已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2(1)判断{1/Sn},、{An}是否是等差数列,(2)
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![已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2 (1)判断{1/Sn},、{An}是否是等差数已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2(1)判断{1/Sn},、{An}是否是等差数列,(2)](/uploads/image/z/766721-65-1.jpg?t=%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%2C%E4%B8%94%E6%BB%A1%E8%B6%B3an%2B2SnSn-1%3D0%28n%E2%89%A52%2Cn%E2%88%88N%2B%29%2Ca1%3D1%2F2+%281%29%E5%88%A4%E6%96%AD%7B1%2FSn%7D%2C%E3%80%81%7BAn%7D%E6%98%AF%E5%90%A6%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%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%2C%E4%B8%94%E6%BB%A1%E8%B6%B3an%2B2SnSn-1%3D0%28n%E2%89%A52%2Cn%E2%88%88N%2B%29%2Ca1%3D1%2F2%281%29%E5%88%A4%E6%96%AD%7B1%2FSn%7D%2C%E3%80%81%7BAn%7D%E6%98%AF%E5%90%A6%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%EF%BC%882%EF%BC%89)
已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2 (1)判断{1/Sn},、{An}是否是等差数已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2(1)判断{1/Sn},、{An}是否是等差数列,(2)
已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2 (1)判断{1/Sn},、{An}是否是等差数
已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2
(1)判断{1/Sn},、{An}是否是等差数列,
(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+2SnSn-1=0(n≥2,n∈N+),a1=1/2 (1)判断{1/Sn},、{An}是否是等差数已知数列{an}的前n项和为Sn,且满足an+2SnSn-1=0(n≥2,n∈N+),a1=1/2(1)判断{1/Sn},、{An}是否是等差数列,(2)
第一问:
由数列公式可得出Sn-Sn-1=an 将此等式带入已知等式
可得出:Sn-Sn-1+2SnSn-1=0 等式同时除以2SnSn-1 即可得出1/Sn为等差数列
第二问:
由于知道1/Sn为公差为2的等差数列 可先求出1/Sn=2n Sn=1/2n
再由公式Sn-Sn-1=an 求出an=1/n(1-2n) an不是等差数列
解 ∵an+2SnSn-1=0 且an=Sn-Sn-1 ∴Sn-Sn-1+2SnSn-1=0 给等式同除SnSn-1 可得1/Sn-1/Sn-1=2 所以{1/Sn}为等差数列 则1/Sn=2n 故1/Sn-1=2(n-1) 所以 an=1/2n-1/2(n-1) 故、{An}不是等差数列