Question

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

Answer 1

(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