設(shè){an}的公差為d.
Sn=na1+n(n-1)d/2、Sn/n=a1+(n-1)(d/2).
所以,數(shù)列{Sn}是首項為a1=2、公差為d/2的等差數(shù)列.
S2009/2009-S2007/2007=2(d/2)=d=2.
an=2+2(n-1)=2n.
Sn=n(2+2n)/2=n(n+1)
數(shù)列{1/Sn}的前n項和為：
Tn=1/S1+1/S2+…+1/Sn
=1/(1*2)+1/(2*3)+…+1/[n(n+1)]
=1-1/2+1/2-1/3+…+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)