早教吧作业答案频道 -->数学-->
已知等差数列,an=2n+1,令Tn=1/S1+1/S2+1/S3+……+1/Sn,求证：Tn＜3/4
题目详情
已知等差数列,an=2n+1,令Tn=1/S1+1/S2+1/S3+……+1/Sn,求证：Tn＜3/4
▼优质解答
答案和解析
由等差数列,an=2n+1得a1=3
所以sn=n(3+2n+1)/2=n(n+2)
所以1/sn=1/[n(n+2)]=1/2*[1/n-1/(n+2)]
因此Tn=1/S1+1/S2+1/S3+……+1/Sn
=1/2*[1-1/3+1/2-1/4+1/3-1/5+.+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2*[1+1/2-1/(n+1)-1/(n+2)]
=1/2*[3/2-1/(n+1)(n+2)]
=3/4-1/[2(n+1)(n+2)]＜3/4
所以Tn＜3/4
所以sn=n(3+2n+1)/2=n(n+2)
所以1/sn=1/[n(n+2)]=1/2*[1/n-1/(n+2)]
因此Tn=1/S1+1/S2+1/S3+……+1/Sn
=1/2*[1-1/3+1/2-1/4+1/3-1/5+.+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2*[1+1/2-1/(n+1)-1/(n+2)]
=1/2*[3/2-1/(n+1)(n+2)]
=3/4-1/[2(n+1)(n+2)]＜3/4
所以Tn＜3/4
看了 已知等差数列,an=2n+1...的网友还看了以下：