Question

已知数列an的首项为a1=3，an与前n项和Sn之间满足2an=Sn*S(n-1) (n>1)

求证｛1/Sn｝是等差数列,并求公差 求数列an的通项公式

Answer 1

2an=sn*s(n-1)
2sn-2s(n-1)=sn*s(n-1)
1/s(n-1)-1/sn=1/2
1/sn-1/s(n-1)=-1/2
｛1/Sn｝是首项为1/3公差为-1/2的等差数列
1/sn=1/3-(n-1)/2
1/sn=(5-3n)/6
sn=6/(5-3n)
an=sn-s(n-1)
an=6/(5-3n)-6/(5-3(n-1))
an=6(1/(5-3n)-1/(8-3n))
an=18/((5-3n)(8-3n))