已知各项均为正数的数列{an},其前n项和为Sn,且满足2Sn=an2+an.求：若数列{1/an2}的前n项和为Tn,

2Sn = an^2 +an
2(Sn - Sn-1) = an^2 +an -(an-1^2 +an-1)
2an = an^2 +an -(an-1^2 +an-1)
an^2 -an -an-1^2 -an-1 =0
因式分解
an^2-an-1^2 -(an + an-1) =0
(an +an-1)(an- an-1 -1)=0
因为 {an} 为正数数列
只能an- an-1 -1=0
an - an-1 = 1, 是等差数列
2S1 = 2a1 = a1^2 +a1
a1(a1-1)=0
a1=1
通项公式 an = 1+(n-1)*1 = n
Tn = 1 + 1/2^2 +1/3^3+...+1/n^2
n>=3时
Tn > 1+ 1/(2*3) + 1/(3*4) +... +1/n(n+1)
= 1 + 1/2 -1/3 +1/3 -1/4 +...+ 1/n -1/(n+1)
= 3/2 -1/(n+1)
又因为n>=3
3/2 -1/(n+1) -[ 3/2 +(1-2n)/2n^2]
= -1/(n+1) -1/2n^2 + 1/n
= 1/n-1/(n+1) -1/2n^2
=1/n(n+1) -1/2n^2
= 1/(n^2+n) -1/(n^2+n^2)
>0
所以
Tn >3/2 -1/(n+1) >3/2 +(1-2n)/2n^2

2Sn=an²+an
2S(n-1)=a(n-1)²+a(n-1)
2an=an²+an-a(n-1)²-a(n-1)
[an+a(n+1)][an-a(n-1)-1]=0
∵各项均为正数
∴an-a(n-1)=1
2S1=a1²+a1
a1=1
an=a1+(n-1)d=n
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