已知等比数列{an}的各项均为正数,且2a1+3a2=1,a32=9a2a6.

2个回答

  • (1)a3^2=9a2a6

    (a2p)^2=9a2(a2p^4)

    a2^2p^2=9a2^2p^4

    ∵此数列各项均为正数∴a2^20,p>0

    两边同时除以a2^2p^2,得9p^2=1,p=1/3

    2a1+3a2=1

    2a1+3*[(1/3)*a1]=1

    2a1+a1=1

    3a1=1

    a1=1/3

    an=a1p^(n-1)=1/3*(1/3)^(n-1)=1/3^n

    (2)bn=log3a1+log3a2+...+log3an

    =log3(a1*a2*...*an)

    =log3[(1/3)*(1/3^2)*...*(1/3^n)]

    =log3[(1/3)^(1+2+...+n)]

    =(1+2+...+n)*log3(1/3)

    =-n(n+1)/2

    1/bn=-2/n(n+1)

    =(-2)*[1/n(n+1)]

    =(-2)*[1/n-1/(n+1)]

    1/b1+1/b2+...+1/bn

    =(-2)*(1-1/2)+(-2)*(1/2-1/3)+...+(-2)*[1/n-1/(n+1)]

    =(-2)*[1-1/2+1/2-1/3+...+1/n-1/(n+1)]

    =(-2)*[1-1/(n+1)]

    =-2n/(n+1)