设各项均为正数的数列{an}和{bn}满足5^an,5^bn,5^an+1成等差数列,Igbn,Iogan+1,Igbn

1个回答

  • 应该是5^An,5^Bn,5^A(n+1)成等比数列

    (5^Bn)^2=5^An×5^A(n+1)

    5^(2Bn)=5^[An+A(n+1)]

    2Bn=An+A(n+1)

    又有lgBn,lgA(n+1),lgB(n+1)成等差数列

    2lgA(n+1)=lgBn+lgB(n+1)

    lg[A(n+1)]^2=lg[Bn×B(n+1)]

    [A(n+1)]^2=Bn×B(n+1)

    A(n+1)=[Bn×B(n+1)]^0.5

    n>=2时,An

    2Bn=An+A(n+1)=[B(n-1)×Bn]^0.5+[Bn×B(n+1)]^0.5

    2Bn^0.5=B(n-1)^0.5+B(n+1)^0.5

    {Bn^0.5}为等差数列

    B1=2 B1^0.5=2^0.5

    B2=A2^2/B1=3^2/2=9/2 B2^0.5=3/2×2^0.5

    公差d=B2-B1=3/2×2^0.5-2^0.5=1/2×2^0.5

    Bn^0.5=B1+(n-1)d=2^0.5+(n-1)×1/2×2^0.5=(n+1)/2×2^0.5

    Bn=(n+1)^2/2

    n>=2时

    An=[B(n-1)×Bn]^0.5=[n^2/2×(n+1)^2/2]^0.5=n(n+1)/2

    A1=1也满足上式