题目

已知数列{an}的前n项和为Sn,且-1,Sn,an+1成等差数列,n属于N*,a1=1,函数f（x）=log3X.
1）求数列{an}的通项公式；
2）设数列{bn}满足bn=（n+3）[f(an)+2]分之1,求数列{bn}的前n项和Tn.

提问时间：2020-10-18

答案

1.由已知2Sn=a(n+1)-1
则2S(n-1)=an-1
故2an=2S(n1+)-2Sn=a(n+1)-an
a(n+1)=3an
所以{an}为公比是3的等比数列
an=a1*3^(n-1)=3^(n-1)
2.bn=1/{(n+3)[f(an)+2)]}=1/(n+3)(n-1+2)=1/(n+1)(n+3)=(1/2)[1/(n+1)-1/(n+3)]
Tn=(1/2)(1/2-1/4)+(1/2)(1/3-1/5)+(1/2)(1/4-1/6)+...+(1/2)[1/(n+1)-1/(n+3)]
=(1/2)[1/2+1/3-1/(n+2)-1/(n+3)]
=5/12-(2n+5)/[2(n+2)(n+3)]