Question

已知数列｛an｝中,首项a1=1/2,an+Sn•S(n-1)=0,求数列｛an｝的通项公式

Answer 1

an +Sn.S(n-1) =0

Sn - S(n-1) +Sn.S(n-1) =0

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

{1/Sn} 是等差数列, d=1

1/Sn -1/S1 =n-1

1/Sn = n+1

Sn = 1/(n+1)

an = Sn - S(n-1)

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

= -1/[n(n+1)]

an = 1/2 ; n=1

= -1/[n(n+1)] ; n=2,3,4,.