Question

数列｛An｝的首项A1=3,通项An 前n项和 Sn之间满足2An=Sn×Sn－1（n≥2,n∈N*) (1) 求证：数

Answer 1

1.

Sn×S(n-1)=2An=2(Sn-S(n-1))

两边除以Sn×S(n-1)

1=2(Sn-S(n-1))/(Sn×S(n-1))

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

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

{1/Sn}是公差为-1/2的等差数列

2.

1/S1=1/A1=1/3

1/Sn=1/3+(n-1)×(-1/2)=(5-3n)/2

Sn=6/(5-3n)

n>=2时

An=Sn-S(n-1)=6/(5-3n)-6/(5-3(n-1))=18/((3n-5)(3n-8))

A1=3