- 有关等差数列前n项和的解答题已知A1=1,Sn=n^2An,求A
- 已知 A1=1 , Sn=n^2An ,求An及Sn?
(过程怎么写)
- 解:
由Sn=n^2An 得Sn=n^2*(Sn-Sn-1) n>=2
整理得Sn/Sn-1=n^2/(n^2-1)
=[n/(n-1)]*[n/(n+1)] n>=2
Sn=[(Sn/Sn-1)*(Sn-1/Sn-2)*...*(S2/S1)]*S1
={[n/(n-1)]*[(n-1)/(n-2)]*...*(2/1)}*{[n/(n+1)]*[(n-1)/n]*...*(2/3)}*1
=2n/(n+1)
所以An=Sn-Sn-1
=2n/(n+1)-2(n-1)/n
=2/[n(n+1)] n>=2
当n=1,A1=1也符合上式,所以
An=2/[n(n+1)] n为自然数