Question

急数学问题求答

已知数列An的前n项和为Sn.且满足an+2Sn*Sn-1=0<n>=2>,a1=1/2,求证1/Sn是等差数列，求通项an的表达试

Answer 1

证明1/Sn等差数列，即证1/Sn - 1/Sn-1 =常数。当n=2时，a2+2(a1+a2)*a1=0，a2=-1/4，所以1/S2-1/S1=2。n>=2时，an=Sn-Sn-1带入an+2Sn*Sn-1=0得Sn-Sn-1+2Sn*Sn-1=0
对此式两边同除以Sn*Sn-1，得1/Sn-1/Sn-1=2，得证

Answer 2

an=Sn-Sn-1
Sn-Sn-1+2*Sn*Sn-1=0
1/Sn -1/Sn-1 =2
所以1/Sn为等差数列
n=2时, a2+2*（a1+a2）*a1=0
a1=1/2，a2=-1/4
S1=1/2; S2=1/4
1/S1=2 1/S2=4
1/Sn=2n
Sn=1/2n
an=Sn-Sn-1=1/2n-(1/2（n-1）)

Answer 3

<1>an+2Sn*Sn-1=0
Sn-Sn-1+2Sn*Sn-1=0
两边同除Sn*Sn-1则1/Sn-1+1/Sn+2=0
1/Sn-1/Sn-1=2所以1/Sn为等差数列
<2>当n=2，1/S1=1/a1=2
1/Sn=2n,则Sn=1/2nSn-1=1/2(n-1)
an=Sn-Sn-1=1/2n-1/2(n-1)=-1/2n（n-1）