设数列{an}与{bn}分别为正项等比数列,Tn与Rn别为数列{lgan}与{lgbn}的前n项和,且Tn/Rn=n/(2n+1),则logb5a5=(b5为对数中的下标,b5中的5又为数列的下标,a5为对数中的上标,a5中的5又为数列的下标)

Question

设数列{an}与{bn}分别为正项等比数列,Tn与Rn别为数列{lgan}与{lgbn}的前n项和,且Tn/Rn=n/(2n+1),则logb5a5=
(b5为对数中的下标,b5中的5又为数列的下标,a5为对数中的上标,a5中的5又为数列的下标)

Answer 1

容易,首先logb5a5=lga5/lgb5,{Tn}和{Rn}分别是{lgan}和{lgbn}的前N项和,且Tn/Rn=n/(2n+1),所以T9/R9=(lga1+lga2+…+lga9)/(lgb1+lgb2+…+lgb9)=[lg(a1*a2*…*a9)]/[lg(b1*b2*…*b9)]=[lg(a5)^5]/[lg(b5)^5]=5lga5/5lgb5=9/19,即logb5a5=9/19