(1) an=sn-s(n-1)
就有sn-s(n-1)+2sn*s(n-1)=0
两边同除以sn*s(n-1)得
1/sn-1/s(n-1)=2 {1/sn}是等差数列
1/sn=1/s1+(n-1)d=2n-1
sn=1/(2n-1)
(2) bn=1/[(2n-1)(2n+1)]=1/2*[1/(2n-1)-1/(2n+1)]
数列{bn}的前n项和Tn=1/2*[1-1/3+1/3-1/5+……+1/(2n-1)-1/(2n+1)]
=1/2*[1-1/(2n+1)]=n/(2n+1)