n=1时,S1=a1=1-4+4=1
n≥2时,
Sn=n²-4n+4 S(n-1)=(n-1)²-4(n-1)+4
Sn-S(n-1)=an=n²-4n+4-(n-1)²+4(n-1)-4=2n-5
n=1时,a1=2-5=-3,与a1=1不符.
数列{an}的通项公式为
an=1 n=1
2n-5 n≥2
n=1时,T3-T1=1/a1+1/a2+1/a3-1/a1=1/a2+1/a3=1/(4-5)+1/(6-5)=-1+1=0
T3-T1≤m/15
m/15≥0
m≥0
n≥2时,
Tn=1/a1+1/a2+...+1/an
T(2n+1)=1/a1+1/a2+...+1/an+1/a(n+1)+1/a(n+2)+...+1/a(2n+1)
T(2n+1)-Tn=1/a(n+1)+1/a(n+2)+...+1/a(2n+1)
[T(2n+3)-T(n+1)]-[T(2n+1)-Tn]
=[1/a(n+2)+1/a(n+3)+...+1/a(2n+1)+1/a(2n+2)+1/a(2n+3)]-[1/a(n+1)+1/a(n+2)+...+1/a(2n+1)]
=1/a(2n+2)+1/a(2n+3)-1/a(n+1)
=1/a(2n+3)-1/(2n+2)