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

设公差d，则
a1+a1+2d=6 a1+10d=21
∴a1=1 d=2
∴an=1+2(n-1)=2n-1 sn=2*(n+1)*n/2-n=n²
∴bn=1/【n*(2n+2)】=1/2*[1/n-1/(n+1)]
故前n项的和：1/2[1-1/(n+1)]=n/(2n+2)

a1+a3=a1+a1+2d=2a1+2d=6
a1+d=3 1
a11=a1+10d=21 2
2式-1式得
9d=18
d=2
a1=1

an=a1+(n-1)d=1+(n-1)*2=2n-1

bn=1/n(an+3)
=1/n(2n+3-1)=1/2n(n+1)=1/2*(1/n-1/(n+1))

Tn=b1+b2+b3+.........+bn
=1/2(1-1/2+1/2-1/3+..........+1/n-1/(n+1))
=1/2(1-1/(n+1))
=n/2(n+1)

∵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)]