已知数列{an}满足条件:(n-1)an+1=(n+1)(an-1)(n∈N*),求an通项公式.（前面的an+1中n+1为底数,后面an-1中n为底数）求过程!~~~~~~求详解条件里还有a2=6 忘记打了

(n-1)a(n+1)=(n+1)[a(n)-1],n=1,2,...
n=1,0=2[a(1)-1], a(1)=1.
n=2,a(3)=3[a(2)-1].
n=3,2a(4)=4[a(3)-1],a(4)=2[a(3)-1]=2[3a(2)-4],
由于a(2)可取任何值，因此{a(n)}的通项公式不存在哈~~~

(n-1)an+1=(n+1)(an-1)可化为an+1/(n+1)=an/(n-1)-1/(n-1),（前面的an+1中n+1为底数）两边同时除以n,可得：an+1/n(n+1)=an/n(n-1)-1/n(n-1),设数列an/n(n-1)为bn,则bn+1=bn-[1/n(n-1)](n>1),将1/n(n-1)裂项为1/(n-1)...