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
问题描述：
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
答
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...