Question

数列中,a1=1,Sn=[S(n-1)]/[2S(n-1)+1] (n>=2)求正｛1/Sn｝是等差数列 求通响

Answer 1

1/Sn=[2S(n-1)+1]/S(n-1)=2+1/S(n-1)

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

所以1/Sn是等差数列

S1=a1=1

1/S1=1

1/Sn的d=2

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

Sn=1/(2n-1)

S(n-1)=1/(2n-3)

则n>=2时

an=Sn-S(n-1)=-2/[(2n-1)(2n-3)]

a1=1也符合

所以an=-2/[(2n-1)(2n-3)]