an是等差数列,前n项和Sn a3=6,S3=12 求和1/S1 +1/S2 +1/S3 +.1/Sn

题目

an是等差数列,前n项和Sn a3=6,S3=12 求和1/S1 +1/S2 +1/S3 +.1/Sn

答案

S3=(a1+a3)*3/2=12
a1=2 , d=2
an=2n
Sn=(a1+an)*n/2
=n(n+1)
1/Sn=1/[n(n+1)]=1/n-1/(n+1)
∴1/S1+1/S2+.1/Sn
=(1-1/2)+(1/2-1/3)+.1/n-1/(n+1)
=1-1/(n+1)