an是等差数列,a2=5,a6=21,设数列1/an的前n项和为Sn,S(2n+1)-Sn≤m/15,求m的最小值

an是等差数列,a2=5,a6=21,设数列1/an的前n项和为Sn,S(2n+1)-Sn≤m/15,求m的最小值
an是等差数列,a2=5,a6=21,设数列1/an的前n项和为Sn,S(2n+1)-Sn≤m/15,求m的最小值

an是等差数列,a2=5,a6=21,设数列1/an的前n项和为Sn,S(2n+1)-Sn≤m/15,求m的最小值
设等差数列{an}公差为d.
a6-a2=4d=21-5=16 d=4
a1=a2-d=5-4=1
an=a1+(n-1)d=1+4(n-1)=4n-3
1/an=1/(4n-3)
S(2n+1)-Sn=a1+a2+...+an+a(n+1)+...+a(2n+1)-(a1+a2+...+an)
=a(n+1)+a(n+2)+...+a(2n+1)
a[(n+1)+1]+a[(n+1)+2]+...+a[2(n+1)+1]-[a(n+1)+a(n+2)+...+a(2n+1)]
=a(n+2)+a(n+3)+...+a(2n+1)+a(2n+2)+a(2n+3)-[a(n+1)+a(n+2)+...+a(2n+1)]
=a(2n+2)+a(2n+3)-a(n+1)
=1/[4(2n+2) -3]+1/[4(2n+3) -3] -1/[4(n+1)-3]
=1/(8n+5)+1/(8n+9)-2/(8n+2)