Question

已知数列{an}前n项和Sn=n^2,令{bn}=1/anan+1,数列{bn}的前n项和为Tn。

求{an}的通项公式an和Tn

Answer 1

Sn=n^2
n>=2时，
an=Sn-S(n-1)=n^2-(n-1)^2=2n-1
n=1时，a1=S1=1也满足上式
bn=1/ana(n+1)
=1/[(2n-1)(2n+1)]
=(1/2)[(2n+1)-(2n-1)]/[(2n-1)(2n+1)]
=(1/2)[1/(2n-1)-1/(2n+1)]
Tn=B1+B2+……+Bn
=(1/2)[(1/1-1/3)+(1/3-1/5)+……+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)