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已知等比数列{An}中,A1=1/3公比q=1/3,设Bn=㏒3A2+...

an=a1*q^(n-1)=3^(-1)*3^(1-n)=3^(-n) bn=㏒3a1+㏒3a2+…+㏒3an =(-1)+(-2)+……+(-n) =-n(n+1)/2

解: 由于{an}为等比数列 则:a5a6=a4a7=a3a8=a2a9=a1a10 又a5a6+a4a7=18 则: 2a5a6=18 a5a6=9 则: log3(a1)+log3(a2)+...+log3(a9)+log3(a10) =log3[a1*a2*a3*...*a10] =log3[(a1a10)*(a2a9)*...*(a5a6)] =log3[9*9*...*9] =log3[9^5] =log3[3^10...

解:㏒a2/3<1等价于:loga2/31时,由不等式loga2/3

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