关于等差数列的计算

题目

关于等差数列的计算
a1=2 d=2 求1/S1+1/S2+1/S3+...+1/Sn

答案

Sn=na1+n(n-1)d/2=n^2+n=n*(n+1)
1/Sn=1/n(n+1)=1/n-1/(n+1)
1/S1+1/S2+1/S3+...+1/Sn
=(1-1/2)+(1/2-1/3)+……+(1/n-1/n+1)
=1-1/(n+1)
=n/(n+1)