设{an}的公差为d.
Sn=na1+n(n-1)d/2、Sn/n=a1+(n-1)(d/2).
所以,数列{Sn}是首项为a1=2、公差为d/2的等差数列.
S2009/2009-S2007/2007=2(d/2)=d=2.
an=2+2(n-1)=2n.
Sn=n(2+2n)/2=n(n+1)
数列{1/Sn}的前n项和为：
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)