sn=n^2+2n
s(n-1)=(n-1)^2+2(n-1)=n^2-1
所以an=sn-s(n-1)=(n^2+2n)-(n^2-1)=2n+1
a1=s1=1+2=3
符合an=2n+1
所以an=2n+1
a1=3,a2=5,a3=7,……
所以Tn=1/3*5+1/5*7+……+1/(2n+1)(2n-1)
=[2/3*5+2/5*7+……+2/(2n+1)(2n-1)]/2
={(1/3-1/5)+(1/5-17)+……+[1/(2n-1)-1/(2n+1)]}/2
=[1/3-1/(2n+1)]/2
=(n-1)/(2n+1)