已知数列an为等差数列且a1=2 a1+a2+a3=12令bn=1/（an+1)(an+3)求bn的前n项和

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)]

S3=3a1+3d=6+3d=12
d=2
bn=1/a(n+1)a(n+3)=1/(2+2n)(2+2(n+2))=1/4(n+1)(n+2)=1/4*[1/(n+1)-1/(n+2)]
S(bn)=1/4*[1/2-1/3+1/3-1/4+……+1/(n+1)-1/(n+2)]
=1/4*[1/2-1/(n+2)]=n/8(n+2)