Question

已知数列an的前n项和为Sn,且满足a1=1/2,an=-2SnSn-1(n≥2).

Answer 1

1.

n>=2时

An=Sn-S(n-1)

Sn-S(n-1)+2SnS(n-1)=0

两边除以SnS(n-1)

1/S(n-1)-1/Sn+2=0

1/Sn-1/S(n-1)=2

{1/Sn}是公差为2的等差数列

2.

1/S1=1/A1=2

1/Sn=2n

Sn=1/(2n)

An=Sn-S(n-1)=1/(2n)-1/(2n-2)=-1/(2n(n-1))

A1=1/2

3.

Bn=2(1-n)An=1/n

(Bn)^2=1/n^2