Question

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

Answer 1

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)