Question

数列a1=3,数列满足2an=Sn乘以S(n-1),求通项公式.

Answer 1

因为an=Sn-S(n-1) n>=2

2an=Sn*S( n-1)

2(Sn-S(n-1))=Sn*S(n-1) 两边除以Sn*S(n-1)

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

所以数列{1/Sn}是以-1/2为公差 1/S1=1/3为首项的等差数列

1/Sn=-n/2 + 5/6

Sn=1/(-n/2 +5/6)

an=Sn-S(n-1)=1/(-n/2 +5/6)- 1/(-n/2+4/3) =6/(5-3n)-6/(8-3n) n>=2

a1=3 不满足an

所以 an =6/(5-3n)-6/(8-3n) n>=2

=3 n=1