Question

已知{an}为等差数列,前n项和为Sn,a3=6,S3=12, an=2n(2)求1/S1+1/S2+…+1/Sn

an=2n 后面多了（2）

Answer 1

an=2n
Sn=2+4+6+……+2n=2*n(n-1)/2=n(n-1)

1/S1+1/S2+…+1/Sn=1/1*2+1/2*3+……+1/n(n-1)
=[1/1-1/2+1/2-1/3+1/3-1/4+……+1/(n-1)-1/n]
=1-1/n=（n-1)/n