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)
问题描述:
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
答
(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)...