已知数列{an}的前n项和为Sn=1/2n^2+1/2n.设Tn=1/a1a2+1/a2a3+1/a3a4+……+1/a

{an} 是等差数列,
a1=S1=1,S2=(1/2)*4+(1/2)*2=3
a2=S2-S1=2
d=1
an=1+(n-1)*1=n
1/[an*a(n+1)]=1/[n(n+1)]=1/n-1/(n+1)
Tn=1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)