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

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

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

sn=a1n+n(n-1)d/2=2n+n(n-1)=n(n+1)
1/sn=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)

由已知得Sn=n（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