是关于数列证明的!如题：{An}中,A1为三分之一,前n项的和除以项数等于（2n-1)乘以An,怎么证明An等于[（2n+1）乘以（2n-1）]分之一?现在是把n等于k+1带进去,怎么才能算出A(k+1)等于[2乘以（k+1)-1]乘以[2乘以（k+1）+1]分之一?

{An},A1=1/3,Sn/n=(2n-1)An,求证:An=1/[（2n+1）（2n-1）]
证明：Sn=n*(2n-1)An
Sn+1=(n+1)*(2n+1)An+1
Sn+1-Sn=(n+1)*(2n+1)An+1-n*(2n-1)An
An+1=(n+1)*(2n+1)An+1-n*(2n-1)An
(2n+3)An+1=(2n-1)An
(2n+3)(2n+1)An+1=(2n-1)(2n+1)An
Bn=(2n-1)(2n+1)An
则Bn+1=Bn=.=B1=1*3/3=1
Bn=(2n-1)(2n+1)An=1
An=1/[(2n-1)(2n+1)]
毕!