已知数列an的前n项和为Sn,且满足an+2Sn·S(n-1)=0(n≥2),a1=1.5

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已知数列an的前n项和为Sn,且满足an+2Sn·S(n-1)=0(n≥2),a1=1.5
⑴求证1/Sn是等差数列
⑵求an的表达式
⑶若bn=2(1-n)·an(n≥2)求证b2·b2+b3·b3+…+bn·bn

(1)an+2Sn·S(n-1)=0(n≥2),又an=Sn-S(n-1)
所以Sn-S(n-1)+2Sn·S(n-1)=0(n≥2)
两边同时除以Sn·S(n-1),得1/S(n-1)-1/sn+2=0
即 1/sn-1/S(n-1)=2
所以1/Sn是公差为2的等差数列.
(2)s1=a1=1.5
1/s1=2/3
公差为2
1/sn=2n-4/3
sn=3/（6n-4),S(n-1)=3/(6n-10)
an=Sn-S(n-1)=3/（6n-4)-3/(6n-10)=-18/[(6n-4)*(6n-10)]=-9/(18n^2-42n+20)
(3)bn=2(1-n)·an=2(1-n)[-9/(18n^2-42n+20)]=-9(1-n)/[(3n-2)(3n-5)]
b2·b2+b3·b3+…+bn·bn

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