1、在等差数列{an}中,Sn=n*n+2n,若{bn}=1/an*a（n+1）.求数列{bn}的前n项和Tn

题目

1、在等差数列{an}中,Sn=n*n+2n,若{bn}=1/an*a（n+1）.求数列{bn}的前n项和Tn

答案

Sn=n*n+2n,则Sn-1=(n-1)^2+2(n-1)=n^2-1
an=Sn-Sn-1=2n+1
bn=1/an*a（n+1）=1/(2n+1)(2n+3)=1/2[1/(2n+1)-1/(2n+3)]
Tn=b1+b2+b3+...+bn
=1/2[1/3 - 1/5 + 1/5- 1/7 + 1/7 - 1/9 +.1/(2n+1)-1/(2n+3)]
=1/2[1/3-1/(2n+3)]
=n/3(2n+3)