已知数列{an}的前n项和为Sn,且满足an+2SnS(n-1)=0(n≥2),a=1/2

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已知数列{an}的前n项和为Sn,且满足an+2SnS(n-1)=0(n≥2),a=1/2
1）证明{1/sn}是等差数列
2）求数列{an}的通项an
3）若bn=2(1-n)an,(n>=2),求证：b2^2+b3^2+...+bn^2

1)
an = Sn - S(n-1) (n>=2)
所以 Sn - S(n-1) + 2SnS(n-1)=0
所以 1/S(n-1) - 1/Sn + 2=0
1/Sn - 1/S(n-1) = 2
所以{1/sn}是等差数列,首项是 1/S1 = 1/a1 = 2,公差是2
2）
由1知
1/Sn = 2 + (n-1)*2 = 2n
所以 Sn = 1/(2n)
an = Sn - S(n-1) = 1/(2n) - 1/(2n - 2) = -1/[2n(n-1)] (n>=2)
n = 1时,a1 = 1/2
3)
bn = 2(1-n)an = 1/n (n>=2)
b2^2+b3^2+...+bn^2 = 1/2^2 + 1/3^2 +…… + 1/n^2

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