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