【数学】数列证明题已知f(x)=x^2-1/2x+1/4,数列{bn}满足:b1=b,b(n+1)=2f(bn)(n∈N+),若1/2<b<1,且{1/bn}的前n项和为Sn,求证:Sn<2/(2b-1)
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![【数学】数列证明题已知f(x)=x^2-1/2x+1/4,数列{bn}满足:b1=b,b(n+1)=2f(bn)(n∈N+),若1/2<b<1,且{1/bn}的前n项和为Sn,求证:Sn<2/(2b-1)](/uploads/image/z/223498-10-8.jpg?t=%E3%80%90%E6%95%B0%E5%AD%A6%E3%80%91%E6%95%B0%E5%88%97%E8%AF%81%E6%98%8E%E9%A2%98%E5%B7%B2%E7%9F%A5f%28x%29%3Dx%5E2-1%2F2x%2B1%2F4%2C%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Ab1%3Db%2Cb%28n%2B1%29%3D2f%28bn%29%28n%E2%88%88N%2B%29%2C%E8%8B%A51%2F2%EF%BC%9Cb%EF%BC%9C1%2C%E4%B8%94%7B1%2Fbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%B1%82%E8%AF%81%EF%BC%9ASn%EF%BC%9C2%2F%282b-1%29)
【数学】数列证明题已知f(x)=x^2-1/2x+1/4,数列{bn}满足:b1=b,b(n+1)=2f(bn)(n∈N+),若1/2<b<1,且{1/bn}的前n项和为Sn,求证:Sn<2/(2b-1)
【数学】数列证明题
已知f(x)=x^2-1/2x+1/4,数列{bn}满足:b1=b,b(n+1)=2f(bn)(n∈N+),若1/2<b<1,且{1/bn}的前n项和为Sn,求证:Sn<2/(2b-1)
【数学】数列证明题已知f(x)=x^2-1/2x+1/4,数列{bn}满足:b1=b,b(n+1)=2f(bn)(n∈N+),若1/2<b<1,且{1/bn}的前n项和为Sn,求证:Sn<2/(2b-1)
变态题目啊
好难好难