- [急求助]高二数学
- 请点击看大图,非常感谢您的解答.
- 解：设等差数列首项为a1，公差为d
Sn=[2a1n+n（n-1）d]/2
S3=[6a1+6d]/2=3a1+3d
S5=[10a1+20d]/2=5a1+10d
S3+S5=8a1+13d=21.............(1)
b3=1/S3=1/(3a1+3d)
a3=a1+2d
b3a3=(a1+2d)/(3a1+3d)=1/2.....(2)
解(1)(2)得: a1=d=1
∴bn=1/Sn=2/[2n+n（n-1）]=2/n(n+1)=2[1/n-1/(n+1)]
b1+b2+......bn=Tn=
=2[(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+....1/(n-1)-1/n+1/n-1/(n+1)
=2[1/1-1/(n+1)]=2n/(n+1)
当n→∞时, n/(n+1)＜1
∴Tn＜2