a>0,a不为0,数列{an}前n项和为Sn,满足[(a^n)-1]/Sn=1-(1/a),令数列{bn},bn=an*lgan,求{bn}的前n项和Tn
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a>0,a不为0,数列{an}前n项和为Sn,满足[(a^n)-1]/Sn=1-(1/a),令数列{bn},bn=an*lgan,求{bn}的前n项和Tn
a>0,a不为0,数列{an}前n项和为Sn,满足[(a^n)-1]/Sn=1-(1/a),令数列{bn},bn=an*lgan,求{bn}的前n项和Tn
a>0,a不为0,数列{an}前n项和为Sn,满足[(a^n)-1]/Sn=1-(1/a),令数列{bn},bn=an*lgan,求{bn}的前n项和Tn
当a不等于1时,
由[(a^n)-1]/Sn=1-(1/a),
得 Sn=a*(a^n-1)/(a-1)……(1)
此时 S(n-1)=a*[a^(n-1)-1]/(a-1)……(2)
(1)-(2)有 an=a[a^n-a^(n-1)]/(a-1)
=a^n (n belongs to N*)
所以 bn=an·lgan
=a^n·lga^n
=n·a^n·lgn
Tn=lga[a+2a^2+3a^3+……na^n]
所以 aTn=lga[a^2+2a^3+……(n-1)a^n+na^(n+1)]
两式相减得 (1-a)Tn=lga·[(a+a^2+a^3……a^n)-na^(n+1)]
=lga·{a(1-a^n)/(1-a)-n·a^(n+1)]
所以 Tn=lga·{a(1-a^n)/(1-a)-n·a^(n+1)]/(1-a) (n belongs to N*)
当a=1时,是不是做不了呢?怀疑你把题目打错了吧!a不为0---改为1吧?
百分之一百 自己算的 没有查任何资料