设公差为d,
a1=2
a1+a2+a3=a1+a1+d+a1+2d=3a1+3d=12
a1+d=4,d=4-2=2,
an=a1+(n-1)*2=2+2(n-1)=2(n+1)
bn=1/（an+1)(an+3)
=1/2*[1/(an+1)-1/(an+3)]
bn的前n项和
Sn=b1+b2+……+bn
=1/2*(1/3-1/5++1/5-1/7+……+1/(an+1)-1/(an+3)]
=1/2*[1/3-1/(an+3))]
=an/6(an+3)
=(n+1)/[3(2n+5)]