在数列{An}中,A1=1,An=(n+1)*Sn-1/(n-1),求An

S(n-1)=(n-1)An/(n+1)
Sn=nA(n+1)/(n+2)
两式相减得
An=nA(n+1)/(n+2)-(n-1)An/(n+1)

A(n+1)=2(n+2)An/(n+1)
A(n+1)/（n+2)=2An/(n+1)
设Bn=An/(n+1),B1=A1/(1+1)=1/2,
B(n+1)=2Bn
Bn=(1/2)2^(n-1)=2^(n-2)
An=(n+1)Bn=(n+1)2^(n-2),

S(n)=nA(n+1)/(n+2)
S(n-1)=(n-1)A(n)/(n+1)
A(n)=S(n)-S(n-1)=nA(n+1)/(n+2)-(n-1)A(n)/(n+1)
整理：A(n+1)=2(n+2)A(n)/(n+1)
于是：A(n)=2(n+1)A(n-1)/n=2(n+1)/n*2n/(n-1)*2(n-1)/(n-2)*…*2*3/2A1
即:A(n)=(n+1)*2^(n-2).