Question

已知数列an是等差数列,前N项和为Sn,a3=3,S3=6

已知数列an是等差数列,前N项和为Sn,a3=3,S3=6，（1）求数列an的通项公式（2）令bn=1/Sn，Tn为数列bn的前N项Tn；并证明Tn＜2

Answer 1

答案见下图

Answer 2

你好
1)S2-a3=a1+a2=3=a1+a1+d=3
a3=a1+d+d=3
a1=1 d=1
an=n
2)Sn=n(1+n)/2
bn=2/(n(1+n))
Tn=b1+b2+b3+……+bn
=2/(1*2)+2/(2*3)+2/(3*4)+……+2/(n*(n+1))
=2(1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1))
=2*n/(n+1)<2

Answer 3

由 Sn=阿(a1+an)/n
得S3=6=(a1+3)/3 得 a1=15 d=-7.5
则an=15+(n-1)*(-7.5)