已知数列｛an}前n项和Sn=n^2+n,令bn=1/an*a(n+1),求数列｛bn}的前n项和Tn

n>=2
an=Sn-S(n-1)=2n
a1=S1=2,也符合
所以an=2n
则bn=1/[2n(2n+2)]
=1/4*1/n(n+1)
=1/4*[(n+1)-n]/n(n+1)
=1/4[(n+1)/n(n+1)-n/n(n+1)]
=1/4[1/n-1/(n+1)]
所以Tn=1/4*[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=1/4[1-1/(n+1)]
=n/(4n+4)