设{an}是正项等比数列,令bn=1/n*(lna1+lna2+……+lnan).求证：{bn}是等差数列

Question

设{an}是正项等比数列,令bn=1/n*(lna1+lna2+……+lnan).求证：{bn}是等差数列

Answer 1

设数列{an}的公比为q>0、a1>0
bn=(1/n)(lna1+lna2+…+lnan)
=(1/n)[lna1+lna1q+lna1q^2+…+lna1q^(n-1)]
=(1/n)lna1^nq^(1+2+…+n-1)
=(1/n)[nlna1+[n(n-1)/2]lnq]
=lna1+[(n-1)/2]lnq
b(n+1)=lna1+(n/2)lnq
b(n+1)-bn=(1/2)lnq(常数).
b1=lna1
所以,数列{bn}是首项为lna1、公差为(1/2)lnq的等差数列.
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