已知等差数列a[n]通项公式为a[n]=n,Sn是an^2和an的等差中项.求证1/S1+1/S2+...+1/Sn
问题描述：
已知等差数列a[n]通项公式为a[n]=n,Sn是an^2和an的等差中项.求证1/S1+1/S2+...+1/Sn
数学人气：294 ℃时间：2020-04-30 19:20:56
优质解答
Sn是an^2和an的等差中项
sn=1/2 乘(n^2+n)
1/sn=2/(n^2+n)=2/(n+1)n=2[1/n-1/(n+1)]
1/S1+1/S2+...+1/Sn1/S1+1/S2+...+1/Sn
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
sn=1/2 乘(n^2+n)
1/sn=2/(n^2+n)=2/(n+1)n=2[1/n-1/(n+1)]
1/S1+1/S2+...+1/Sn1/S1+1/S2+...+1/Sn
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
我来回答
类似推荐
答
Sn是an^2和an的等差中项
sn=1/2 乘(n^2+n)
1/sn=2/(n^2+n)=2/(n+1)n=2[1/n-1/(n+1)]
1/S1+1/S2+...+1/Sn1/S1+1/S2+...+1/Sn
=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]
=2[1-1/(n+1)]