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已知不等式1n+1+1n+2+1n+3+…+12n>112loga(a?1)+23...

解答:证明:设f(n)=1n+1+1n+2+…+12n(n∈N,n≥2),∴f(n+1)?f(n)=[1n+2+1n+3+…+12(n+1)]?(1n+1+1n+2+…+12n)=12n+1+12n+2?1n+1=1(2n+1)(2n+2)>0,∴f(n)是关于n(n∈N,n≥2)的递增函数,∴f(n)≥f(2)=712.要使原不等式成立,只需:112loga(a?1...

观察数列分子为以0为首项,2为公差的等差数列,分母是以1为首项,2为公差的等差数列,故可得数列的通项公式an=2(n?1)2n?1(n∈Z*),故选C

∵Sn=2an+1,∴Sn=2(Sn+1-Sn),化为Sn+1Sn=32,∴数列{Sn}是等比数列,首项是1∴Sn=(32)n?1.故选:B.

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