等差数列中,a1+a3=6,a11=21设bn=1/n(an+3)求前n项和

∵a1+a3=6
∴2a1+2d=6……①
又∵a11=a1+10d=21……②
∴联立①②式得：a1=1,d=2
∴an=a1+(n-1)d=2n-1
∴bn=1/[n(an+3)]
=1/[n(2n-1+3)]
=1/[2n(n+1)]
=1/2*[1/n-1/(n+1)]
∴前n项和为：
Tn=b1+b2+……+bn
=1/2*(1-1/2)+1/2*(1/2-1/3)+……+1/2*[1/n-1/(n+1)]
=1/2*[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=1/2*[1-1/(n+1)]
=1/2*[n/(n+1)]
=n/[2(n+1)]