Question

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,求数列{bn}的前n

Answer 1

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)