已知{an}是各项为不同正数的等差数列,lga1,lga2,lga4成等差数列.又bn=1[a(2^n)]证明{bn}为等比数列

问题描述:

已知{an}是各项为不同正数的等差数列,lga1,lga2,lga4成等差数列.又bn=1[a(2^n)]证明{bn}为等比数列

证明:设{an}中首项为a1,公差为d.
∵lga1,lga2,lga4成等差数列∴2lga2=lga1+lga4∴a22=a1?a4.
即(a1+d)2=a1(a1+3d)∴d=0或d=a1.
当d=0时,an=a1,bn=1a2n=1a1,∴bn+1bn=1,∴{bn}为等比数列;
当d=a1时,an=na1,bn=1a2n=12na1,∴bn+1bn=12,∴{bn}为等比数列.
综上可知{bn}为等比数列.
打字不易,