已知｛an｝是各项都为正数的数列,其前N项和为Sn,且满足2anSn-an^2=1设bn=2/(4Sn^4-1)求{bn}前n项和为Tn并求证Tn

Question

已知｛an｝是各项都为正数的数列,其前N项和为Sn,且满足2anSn-an^2=1
设bn=2/(4Sn^4-1)求{bn}前n项和为Tn并求证Tn

Answer 1

n=1时,2a1S1-a1²=2S1²-S1²=S1²=1
数列各项均为正,a1>0,S1>0 S1=1
n≥2时,
2[Sn-S(n-1)]Sn-[Sn-S(n-1)]²=1,整理,得
Sn²-S(n-1)²=1
S1²=a1²=1²=1,数列{Sn²}是以1为首项,1为公差的等差数列
Sn²=1+1×(n-1)=n
bn=2/(4Sn⁴-1)=2/(4n²-1)=2/[(2n+1)(2n-1)]=1/(2n-1) -1/(2n+1)=1/(2n-1) -1/[2(n+1)-1]
Tn=b1+b2+...+bn
=1/(2×1-1)-1/(2×2-1)+1/(2×2-1)+1/(2×3-1)+...+1/(2n-1)-1/[2(n+1)-1]
=1 -1/(2n+1)
=2n/(2n+1)
Tn=2n/(2n+1)=(2n+1-1)/(2n+1)=1 -1/(2n+1)
n>0 1/(2n+1)>0 1-1/(2n+1)