知等差数列{an}的前n项和Sn=n^2-n,设bn=1/Sn+2n,求(1)数列{an}de 通项公式;

问题描述:

知等差数列{an}的前n项和Sn=n^2-n,设bn=1/Sn+2n,求(1)数列{an}de 通项公式;
;(2)求数列{bn}的前n项和Tn

1、当n=1时,a1=S1=0;当n≥2时,an=Sn-S(n-1)=[n²-n]-[(n-1)²-(n-1)]=2n-2,则:an=2n-2(n∈N*)2、bn=1/(Sn+2n)=1/[n(n+1)]=(1/n)-[1/(n+1)],则{bn}的前n项和Tn=[1/1-1/2]+[1/2-1/3]+…+[...