数列{an}各项均为正数，其前n项和为sn，且满足2ansn-an2=1,设bn=2/4sn4-1
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2anSn-(an)2=1
n=1, a1=1
2[Sn-S(n-1)]Sn-[Sn-S(n-1)]^2=1
(Sn)^2-[S(n-1)]^2 =1
{(Sn)^2} 是等差数列, d=1
(Sn)^2-(S1)^2 = n-1
(Sn)^2 = n
bn=2/[4(Sn)^4-1]
= 2/(4n^2-1)
= 1/(2n-1) - 1/(2n+1)
Tn = b1+b2+...+bn
= 1 - 1/(2n+1)
= 2n/(2n+1)
n=1, a1=1
2[Sn-S(n-1)]Sn-[Sn-S(n-1)]^2=1
(Sn)^2-[S(n-1)]^2 =1
{(Sn)^2} 是等差数列, d=1
(Sn)^2-(S1)^2 = n-1
(Sn)^2 = n
bn=2/[4(Sn)^4-1]
= 2/(4n^2-1)
= 1/(2n-1) - 1/(2n+1)
Tn = b1+b2+...+bn
= 1 - 1/(2n+1)
= 2n/(2n+1)
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