Question

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

Answer 1

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)